Question

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?

\(\sum\limits_{n = 0}^\infty {{{( - 1)}^{n - 1}}n{e^{ - n}}{\rm{ (}}|{\rm{error}}|{\rm{ }} < 0.01)} \)

Short answer

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

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 b_nbe 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