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Question

\sum _{n=1}^{+\infty}\frac{n}{2^{n}}
Determine the convergence or divergence

  • Determine the convergence or divergence using the nth Term Test

  • Determine the convergence or divergence using the Ratio Test

  • Determine the convergence or divergence using the Integral Test

\textrm{Inconclusive}
Evaluate
\sum _{n=1}^{+\infty}\frac{n}{2^{n}}
Find the limit
\lim _{n\rightarrow +\infty}\left(\frac{n}{2^{n}}\right)
Use the L'Hopital's rule
\lim _{n\rightarrow +\infty}\left(\frac{\frac{d}{dn}\left(n\right)}{\frac{d}{dn}\left(2^{n}\right)}\right)
\text{Use }\frac{d}{dx} x^{n}=n x^{n-1}\text{ to find derivative}
\lim _{n\rightarrow +\infty}\left(\frac{1}{\frac{d}{dn}\left(2^{n}\right)}\right)
\text{Use }\frac{d}{dx} a^{x}=\ln(a)a^{x}\text{ to find derivative}
\lim _{n\rightarrow +\infty}\left(\frac{1}{\ln{\left(2\right)}\times 2^{n}}\right)
Rewrite the expression
\frac{1}{\lim _{n\rightarrow +\infty}\left(\ln{\left(2\right)}\times 2^{n}\right)}
Calculate
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Evaluate
\lim _{n\rightarrow +\infty}\left(\ln{\left(2\right)}\times 2^{n}\right)
Rewrite the expression
\ln{\left(2\right)}\times \lim _{n\rightarrow +\infty}\left(2^{n}\right)
Calculate
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Evaluate
\lim _{n\rightarrow +\infty}\left(2^{n}\right)
Rewrite the expression
2^{\lim _{n\rightarrow +\infty}\left(n\right)}
Calculate
2^{+\infty}
Calculate
+\infty
\ln{\left(2\right)}\times \left(+\infty\right)
Simplify
+\infty
\frac{1}{+\infty}
Divide the terms
0
Solution
\textrm{Inconclusive}
\text{Converges}
Evaluate
\sum _{n=1}^{+\infty}\frac{n}{2^{n}}
Find the limit
\lim _{n\rightarrow +\infty}\left(\left|\frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^{n}}}\right|\right)
Simplify
\lim _{n\rightarrow +\infty}\left(\left|\frac{n+1}{2n}\right|\right)
Remove the absolute value bars
\lim _{n\rightarrow +\infty}\left(\frac{n+1}{2n}\right)
Rewrite the expression
\lim _{n\rightarrow +\infty}\left(\frac{n}{n}\times \frac{1+\frac{1}{n}}{2}\right)
Simplify
\lim _{n\rightarrow +\infty}\left(\frac{1+\frac{1}{n}}{2}\right)
Rewrite the expression
\frac{\lim _{n\rightarrow +\infty}\left(1+\frac{1}{n}\right)}{2}
Rewrite the expression
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Evaluate
\lim _{n\rightarrow +\infty}\left(1+\frac{1}{n}\right)
Rewrite the expression
\lim _{n\rightarrow +\infty}\left(1\right)+\lim _{n\rightarrow +\infty}\left(\frac{1}{n}\right)
Calculate
1+\lim _{n\rightarrow +\infty}\left(\frac{1}{n}\right)
Calculate
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Evaluate
\lim _{n\rightarrow +\infty}\left(\frac{1}{n}\right)
Rewrite the expression
\frac{\lim _{n\rightarrow +\infty}\left(1\right)}{\lim _{n\rightarrow +\infty}\left(n\right)}
Calculate
\frac{1}{\lim _{n\rightarrow +\infty}\left(n\right)}
Calculate
\frac{1}{+\infty}
Calculate
0
1+0
Calculate
1
\frac{1}{2}
Solution
\text{Converges}
\text{Converges}
Evaluate
\sum _{n=1}^{+\infty}\frac{n}{2^{n}}
\text{Write the n-th term }\frac{n}{2^{n}}\text{ as a function}
f\left(x\right)=\frac{x}{2^{x}}
Rewrite the expression
\frac{d}{dx}\left(\frac{x}{2^{x}}\right)
\text{Use differentiation rule }\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{\frac{d}{dx}(f(x))\times g(x)-f(x)\times \frac{d}{dx}(g\left(x)\right)}{\left(g(x)\right)^2}
\frac{\frac{d}{dx}\left(x\right)\times 2^{x}-x\times \frac{d}{dx}\left(2^{x}\right)}{\left(2^{x}\right)^{2}}
\text{Use }\frac{d}{dx} x^{n}=n x^{n-1}\text{ to find derivative}
\frac{1\times 2^{x}-x\times \frac{d}{dx}\left(2^{x}\right)}{\left(2^{x}\right)^{2}}
\text{Use }\frac{d}{dx} a^{x}=\ln(a)a^{x}\text{ to find derivative}
\frac{1\times 2^{x}-x\ln{\left(2\right)}\times 2^{x}}{\left(2^{x}\right)^{2}}
Any expression multiplied by 1 remains the same
\frac{2^{x}-x\ln{\left(2\right)}\times 2^{x}}{\left(2^{x}\right)^{2}}
Use the commutative property to reorder the terms
\frac{2^{x}-\ln{\left(2\right)}\times x\times 2^{x}}{\left(2^{x}\right)^{2}}
Simplify
\frac{2^{x}-\ln{\left(2\right)}\times x\times 2^{x}}{2^{2x}}
Factor the expression
\frac{2^{x}\left(1-\ln{\left(2\right)}\times x\right)}{2^{2x}}
Reduce the fraction
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Calculate
\frac{2^{x}}{2^{2x}}
\text{Use the product rule }\frac{a^{n}}{a^{m}}=a^{n-m}\text{ to simplify the expression}
\frac{1}{2^{2x-x}}
Subtract the terms
\frac{1}{2^{x}}
\frac{1-\ln{\left(2\right)}\times x}{2^{x}}
Rewrite the expression
\frac{1-\ln{\left(2\right)}\times x}{2^{x}}<0
\text{Separate the inequality into }2\text{ possible cases}
\begin{align}&\left\{ \begin{array}{l}1-\ln{\left(2\right)}\times x>0\\2^{x}<0\end{array}\right.\\&\left\{ \begin{array}{l}1-\ln{\left(2\right)}\times x<0\\2^{x}>0\end{array}\right.\end{align}
Solve the inequality
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Evaluate
1-\ln{\left(2\right)}\times x>0
Move the constant to the right side
-\ln{\left(2\right)}\times x>0-1
Removing 0 doesn't change the value,so remove it from the expression
-\ln{\left(2\right)}\times x>-1
Change the signs on both sides of the inequality and flip the inequality sign
\ln{\left(2\right)}\times x<1
Divide both sides
\frac{\ln{\left(2\right)}\times x}{\ln{\left(2\right)}}<\frac{1}{\ln{\left(2\right)}}
Divide the numbers
x<\frac{1}{\ln{\left(2\right)}}
\begin{align}&\left\{ \begin{array}{l}x<\frac{1}{\ln{\left(2\right)}}\\2^{x}<0\end{array}\right.\\&\left\{ \begin{array}{l}1-\ln{\left(2\right)}\times x<0\\2^{x}>0\end{array}\right.\end{align}
\text{Since the left-hand side is always positive,and the right-hand side is always 0,the statement is false for any value of }x
\begin{align}&\left\{ \begin{array}{l}x<\frac{1}{\ln{\left(2\right)}}\\x \notin \mathbb{R}\end{array}\right.\\&\left\{ \begin{array}{l}1-\ln{\left(2\right)}\times x<0\\2^{x}>0\end{array}\right.\end{align}
Solve the inequality
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Evaluate
1-\ln{\left(2\right)}\times x<0
Move the constant to the right side
-\ln{\left(2\right)}\times x<0-1
Removing 0 doesn't change the value,so remove it from the expression
-\ln{\left(2\right)}\times x<-1
Change the signs on both sides of the inequality and flip the inequality sign
\ln{\left(2\right)}\times x>1
Divide both sides
\frac{\ln{\left(2\right)}\times x}{\ln{\left(2\right)}}>\frac{1}{\ln{\left(2\right)}}
Divide the numbers
x>\frac{1}{\ln{\left(2\right)}}
\begin{align}&\left\{ \begin{array}{l}x<\frac{1}{\ln{\left(2\right)}}\\x \notin \mathbb{R}\end{array}\right.\\&\left\{ \begin{array}{l}x>\frac{1}{\ln{\left(2\right)}}\\2^{x}>0\end{array}\right.\end{align}
\text{Since the left-hand side is always positive,and the right-hand side is always 0,the statement is true for any value of }x
\begin{align}&\left\{ \begin{array}{l}x<\frac{1}{\ln{\left(2\right)}}\\x \notin \mathbb{R}\end{array}\right.\\&\left\{ \begin{array}{l}x>\frac{1}{\ln{\left(2\right)}}\\x \in \mathbb{R}\end{array}\right.\end{align}
Find the intersection
\begin{align}&x \notin \mathbb{R}\\&\left\{ \begin{array}{l}x>\frac{1}{\ln{\left(2\right)}}\\x \in \mathbb{R}\end{array}\right.\end{align}
Find the intersection
\begin{align}&x \notin \mathbb{R}\\&x>\frac{1}{\ln{\left(2\right)}}\end{align}
Find the union
x>\frac{1}{\ln{\left(2\right)}}
\text{Since the function }f(x)=\frac{x}{2^{x}}\text{ is positive,continuous,and decreasing for }x \ge 1\text{,use the Integral Test}
\int _{1}^{+\infty}\frac{x}{2^{x}} dx
By definition,rewrite the improper integral using one-sided limit and a definite integral
\lim _{a\rightarrow +\infty}\left(\int _{1}^{a}\frac{x}{2^{x}} dx\right)
Evaluate the integral
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Evaluate
\int _{1}^{a}\frac{x}{2^{x}} dx
Evaluate the integral
\int \frac{x}{2^{x}} dx
Evaluate the integral
\int 2^{-x}x dx
Prepare for integration by parts
\begin{align}&u=x\\&dv=2^{-x}dx\end{align}
Calculate the derivative
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Calculate the derivative
u=x
Evaluate the derivative
du=x^{\prime}dx
Evaluate the derivative
du=1dx
Simplify the expression
du=dx
\begin{align}&du=dx\\&dv=2^{-x}dx\end{align}
Evaluate the integral
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Evaluate the integral
dv=2^{-x}dx
Evaluate the integral
\int 1 dv=\int 2^{-x} dx
Evaluate the integral
v=-\frac{1}{\ln{\left(2\right)}\times 2^{x}}
\begin{align}&du=dx\\&v=-\frac{1}{\ln{\left(2\right)}\times 2^{x}}\end{align}
\text{Substitute }u=x、v=-\frac{1}{\ln{\left(2\right)}\times 2^{x}}、du=dx、dv=2^{-x}dx\text{ for }\int u dv = uv - \int v du
x\left(-\frac{1}{\ln{\left(2\right)}\times 2^{x}}\right)-\int 1\times \left(-\frac{1}{\ln{\left(2\right)}\times 2^{x}}\right) dx
Calculate
-\frac{x}{\ln{\left(2\right)}\times 2^{x}}-\int -\frac{1}{\ln{\left(2\right)}\times 2^{x}} dx
\text{Use the property of integral }\int kf(x) dx=k\int f(x) dx
-\frac{x}{\ln{\left(2\right)}\times 2^{x}}+\int \frac{1}{\ln{\left(2\right)}\times 2^{x}} dx
Evaluate the integral
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Evaluate the integral
\int \frac{1}{\ln{\left(2\right)}\times 2^{x}} dx
Rewrite the expression
\int \frac{1}{\ln{\left(2\right)}}\times 2^{-x} dx
\text{Use the property of integral }\int kf(x) dx=k\int f(x) dx
\frac{1}{\ln{\left(2\right)}}\times \int 2^{-x} dx
\text{Use the substitution }dx=-1 dt\text{ to transform the integral}
\frac{1}{\ln{\left(2\right)}}\times \int 2^{-x}\left(-1\right) dt
Simplify
\frac{1}{\ln{\left(2\right)}}\times \int -2^{-x} dt
\text{Use the substitution }t=-x\text{ to transform the integral}
\frac{1}{\ln{\left(2\right)}}\times \int -2^{t} dt
\text{Use the property of integral }\int kf(x) dx=k\int f(x) dx
\frac{1}{\ln{\left(2\right)}}\times \left(-1\right)\times \int 2^{t} dt
Multiplying or dividing an odd number of negative terms equals a negative
-\frac{1}{\ln{\left(2\right)}}\times \int 2^{t} dt
\text{Use the property of integral }\int a^{x} dx = \frac{ a^{x}}{\ln a}
-\frac{1}{\ln{\left(2\right)}}\times \frac{2^{t}}{\ln{\left(2\right)}}
Multiply the terms
-\frac{2^{t}}{\ln{\left(2\right)}\times \ln{\left(2\right)}}
Multiply the terms
-\frac{2^{t}}{\left(\ln{\left(2\right)}\right)^{2}}
Substitute back
-\frac{2^{-x}}{\left(\ln{\left(2\right)}\right)^{2}}
Divide the terms
-\frac{1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{x}}
-\frac{x}{\ln{\left(2\right)}\times 2^{x}}-\frac{1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{x}}
Reduce fractions to a common denominator
-\frac{x\ln{\left(2\right)}}{\ln{\left(2\right)}\times 2^{x}\ln{\left(2\right)}}-\frac{1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{x}}
Multiply the numbers
-\frac{x\ln{\left(2\right)}}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{x}}-\frac{1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{x}}
Write all numerators above the common denominator
\frac{-x\ln{\left(2\right)}-1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{x}}
Use the commutative property to reorder the terms
\frac{-\ln{\left(2\right)}\times x-1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{x}}
\text{Use }\frac{-a}{b}=\frac{a}{-b}=-\frac{a}{b}\text{ to rewrite the fraction}
-\frac{\ln{\left(2\right)}\times x+1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{x}}
Return the limits
\left(-\frac{\ln{\left(2\right)}\times x+1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{x}}\right)\bigg |_{1}^{a}
Substitute the values into formula
-\frac{\ln{\left(2\right)}\times a+1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}}-\left(-\frac{\ln{\left(2\right)}\times 1+1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{1}}\right)
Any expression multiplied by 1 remains the same
-\frac{\ln{\left(2\right)}\times a+1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}}-\left(-\frac{\ln{\left(2\right)}+1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{1}}\right)
Evaluate the power
-\frac{\ln{\left(2\right)}\times a+1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}}-\left(-\frac{\ln{\left(2\right)}+1}{\left(\ln{\left(2\right)}\right)^{2}\times 2}\right)
Multiply the terms
-\frac{\ln{\left(2\right)}\times a+1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}}-\left(-\frac{\ln{\left(2\right)}+1}{2\left(\ln{\left(2\right)}\right)^{2}}\right)
Subtract the terms
-\frac{\ln{\left(2\right)}\times a+1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}}+\frac{\ln{\left(2\right)}+1}{2\left(\ln{\left(2\right)}\right)^{2}}
\lim _{a\rightarrow +\infty}\left(-\frac{\ln{\left(2\right)}\times a+1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}}+\frac{\ln{\left(2\right)}+1}{2\left(\ln{\left(2\right)}\right)^{2}}\right)
Rewrite the expression
\lim _{a\rightarrow +\infty}\left(-\frac{\ln{\left(2\right)}\times a+1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}}\right)+\lim _{a\rightarrow +\infty}\left(\frac{\ln{\left(2\right)}+1}{2\left(\ln{\left(2\right)}\right)^{2}}\right)
Calculate
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Evaluate
\lim _{a\rightarrow +\infty}\left(-\frac{\ln{\left(2\right)}\times a+1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}}\right)
\text{Use }\frac{-a}{b}=\frac{a}{-b}=-\frac{a}{b}\text{ to rewrite the fraction}
\lim _{a\rightarrow +\infty}\left(\frac{-\ln{\left(2\right)}\times a-1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}}\right)
Use the L'Hopital's rule
\lim _{a\rightarrow +\infty}\left(\frac{\frac{d}{da}\left(-\ln{\left(2\right)}\times a-1\right)}{\frac{d}{da}\left(\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}\right)}\right)
Find the derivative
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Evaluate
\frac{d}{da}\left(-\ln{\left(2\right)}\times a-1\right)
\text{Use differentiation rule }\frac{d}{dx}\left(f\left(x\right)\pm g\left(x\right)\right)=\frac{d}{dx}\left(f\left(x\right)\right)\pm \frac{d}{dx}(g(x))
-\frac{d}{da}\left(\ln{\left(2\right)}\times a\right)-\frac{d}{da}\left(1\right)
Calculate
-\ln{\left(2\right)}-\frac{d}{da}\left(1\right)
\text{Use }\frac{d}{dx}(c)=0\text{ to find derivative}
-\ln{\left(2\right)}-0
Removing 0 doesn't change the value,so remove it from the expression
-\ln{\left(2\right)}
\lim _{a\rightarrow +\infty}\left(\frac{-\ln{\left(2\right)}}{\frac{d}{da}\left(\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}\right)}\right)
Find the derivative
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Evaluate
\frac{d}{da}\left(\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}\right)
Simplify
\left(\ln{\left(2\right)}\right)^{2}\times \frac{d}{da}\left(2^{a}\right)
\text{Use }\frac{d}{dx} a^{x}=\ln(a)a^{x}\text{ to find derivative}
\left(\ln{\left(2\right)}\right)^{2}\times \ln{\left(2\right)}\times 2^{a}
Multiply the terms
\left(\ln{\left(2\right)}\right)^{3}\times 2^{a}
\lim _{a\rightarrow +\infty}\left(\frac{-\ln{\left(2\right)}}{\left(\ln{\left(2\right)}\right)^{3}\times 2^{a}}\right)
Simplify the expression
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Evaluate
\frac{-\ln{\left(2\right)}}{\left(\ln{\left(2\right)}\right)^{3}\times 2^{a}}
\text{Use }\frac{-a}{b}=\frac{a}{-b}=-\frac{a}{b}\text{ to rewrite the fraction}
-\frac{\ln{\left(2\right)}}{\left(\ln{\left(2\right)}\right)^{3}\times 2^{a}}
Reduce the fraction
-\frac{1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}}
Rearrange the terms
\frac{-1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}}
\lim _{a\rightarrow +\infty}\left(\frac{-1}{\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}}\right)
Rewrite the expression
\frac{-1}{\lim _{a\rightarrow +\infty}\left(\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}\right)}
Calculate
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Evaluate
\lim _{a\rightarrow +\infty}\left(\left(\ln{\left(2\right)}\right)^{2}\times 2^{a}\right)
Rewrite the expression
\left(\ln{\left(2\right)}\right)^{2}\times \lim _{a\rightarrow +\infty}\left(2^{a}\right)
Calculate
\left(\ln{\left(2\right)}\right)^{2}\left(+\infty\right)
Simplify
+\infty
\frac{-1}{+\infty}
Divide the terms
0
0+\lim _{a\rightarrow +\infty}\left(\frac{\ln{\left(2\right)}+1}{2\left(\ln{\left(2\right)}\right)^{2}}\right)
Calculate
0+\frac{\ln{\left(2\right)}+1}{2\left(\ln{\left(2\right)}\right)^{2}}
Calculate
\frac{\ln{\left(2\right)}+1}{2\left(\ln{\left(2\right)}\right)^{2}}
Solution
\text{Converges}
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