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Found in: Page 463

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy?$$\sum\limits_{n = 0}^\infty {{{( - 1)}^{n - 1}}n{e^{ - n}}{\rm{ (}}|{\rm{error}}|{\rm{ }} < 0.01)}$$

Yes, the series is convergent and the number of terms we need to add atleast $$6$$ terms.

See the step by step solution

## Alternating series test:

$$n{e^{ - n}}$$ will always be positive for positive $$n$$ and $${( - 1)^{n - 1}}$$ is positive on odd terms and negative on even terms.

Also, the terms decrease in size. For any $$k$$,consider the size difference between the $${(k + 1)^{{\rm{th}}}}$$ and $${k^{{\rm{th}}}}$$ term

\begin{aligned}(k + 1){e^{ - (n + 1)}} - n{e^{ - n}} &= \frac{{(n + 1)}}{{{e^{n + 1}}}} - \frac{{ - n}}{{{e^n}}}\\ &= \frac{{{e^n}(n + 1) - n{e^{n + 1}}}}{{{e^{2n + 1}}}}\\ &= \frac{{{e^n}\left( {n(1 - e) + 1} \right)}}{{{e^{2n + 1}}}}\end{aligned}

$$1 - e < - 1$$ and $$n \ge 1$$

So, \begin{aligned}n(1 - e) < - 1\\n(1 - e) + 1 < 0\end{aligned}

So the terms are decreasing in size .

## Checking the limit

$$\mathop {\lim }\limits_{n \to \infty } {\rm{ }}n{e^{ - n}} = \mathop {\lim }\limits_{n \to \infty } {\rm{ }}\frac{n}{{{e^n}}}$$

If $$\mathop {\lim }\limits_{n \to \infty } {\rm{ }}\frac{n}{{{e^n}}}$$ exists,

Let $$x = n$$, then $$\mathop {\lim }\limits_{x \to \infty } {\rm{ }}\frac{x}{{{e^x}}} = \mathop {\lim }\limits_{n \to \infty } {\rm{ }}\frac{n}{{{e^n}}}$$

Then \begin{aligned} \mathop {\lim }\limits_{x \to \infty } {\rm{ }}\frac{x}{{{e^x}}} &= \mathop {\lim }\limits_{n \to \infty } {\rm{ }}\frac{1}{{{e^x}}}\\\mathop {\lim }\limits_{x \to \infty } {\rm{ }}\frac{x}{{{e^x}}} &= 0\end{aligned}

Therefore, $$\mathop {\lim }\limits_{n \to \infty } {\rm{ }}\frac{n}{{{e^n}}} = 0$$

So, the series is converging.

Let bn be the sequence $$n{e^{ - n}}$$, $$S = \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}n{e^{ - n}}}$$

Then $$|S - {S_n}| \le {b_{n + 1}}$$

We want $$n$$ with $${b_{n + 1}} < 0.01$$

So that

\begin{aligned} (n + 1){e^{ - (n + 1)}} < 0.01\\\frac{{n + 1}}{{0.01}} < {e^{n + 1}}\end{aligned}

Note that $$\frac{{{e^6}}}{6} = 67.2 < 100$$, while $$\frac{{{e^7}}}{7} = 156.6 < 100$$

So, $$6$$ is the least number that works for $$n + 1$$

Therefore, we need to add atleast $$6$$ terms