Find the sum of the following series:
(a) 
(b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)}
| Background Information:
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| Recall:
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- 1. For a geometric series Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^{\infty} ar^n}
with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |r|<1,}
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- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^{\infty} ar^n=\frac{a}{1-r}.}
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- 2. For a telescoping series, we find the sum by first looking at the partial sum Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_k}
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- and then calculate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{k\rightarrow\infty} s_k.}
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Solution:
(a)
| Step 1:
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| First, we write
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- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\sum_{n=0}^{\infty} (-2)^n e^{-n}} & = & \displaystyle{\sum_{n=0}^{\infty} \frac{(-2)^n}{e^n}}\\ &&\\ & = & \displaystyle{\sum_{n=0}^{\infty} \bigg(\frac{-2}{e}\bigg)^n.}\\ \end{array}}
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| Step 2:
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| Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2<e,~\bigg|-\frac{2}{e}\bigg|<1.}
So,
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- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\sum_{n=0}^{\infty} (-2)^ne^{-n}} & = & \displaystyle{\frac{1}{1+\frac{2}{e}}}\\ &&\\ & = & \displaystyle{\frac{1}{\frac{e+2}{e}}}\\ &&\\ & = & \displaystyle{\frac{e}{e+2}.}\\ \end{array}}
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(b)
| Step 1:
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| This is a telescoping series. First, we find the partial sum of this series.
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| Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_k=\sum_{n=1}^k \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).}
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| Then,
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- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_k=\frac{1}{2}-\frac{1}{2^{k+1}}.}
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| Step 2:
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| Thus,
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- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\sum_{n=1}^{\infty}\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)} & = & \displaystyle{\lim_{k\rightarrow \infty} s_k}\\ &&\\ & = & \displaystyle{\lim_{k\rightarrow \infty}\frac{1}{2}-\frac{1}{2^{k+1}}}\\ &&\\ & = & \displaystyle{\frac{1}{2}.}\\ \end{array}}
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| Final Answer:
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| (a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{e}{e+2}}
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| (b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}}
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