Question

How do you test for convergence of $\sum n{e^{ - n}}$ from $n = [1,\infty )$?

Answer 1

In the above given summation of series, first we need to check whether it is convergent or divergent. Being convergent means we should get a finite value, by finding the range of intervals of r in the geometric series, so further it can be decided whether it is divergent or convergent.

Complete step by step solution:
Convergence should always result in a finite value of summation of series. If it is not finite the series becomes divergent. In the given summation of the series, we get to know that it is a geometric series. It generally looks like the below format:

$\sum\limits_{n = k}^\infty {a{r^n}}$
where $a$is the first term in series and $r$ is the common ratio between the terms.

So, we need to simplify the summation of series to a reduced form.

$$\sum\limits_{n = 1}^\infty {n{e^{ - n}}} \\\ \\\$$

We will apply the ratio test on it. It is generally written as:

$\mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = L$
If $L < 1$,then $\sum\limits_{n = 1}^\infty {{a_n}}$ is absolutely convergent.
If $L > 1$,then $\sum\limits_{n = 1}^\infty {{a_n}}$ is divergent.

Further by substituting n+1 in place of n in numerator and remaining same in denominator we get,

}\limits_{x \to \infty } \dfrac{{(n + 1){e^{ - (n + 1)}}}}{{n{e^{ - n}}}}$$ Now by splitting the exponential power we write it as,

\mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim
}\limits_{x \to \infty } \dfrac{{n + 1}}{n}.\dfrac{{{e^{ - n}}{e^{ - 1}}}}{{{e^{ - n}}}} \\
\mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop
{\dfrac{1}{e}\lim }\limits_{x \to \infty } \dfrac{{n + 1}}{n} \\

By substituting the limit, we get $$\dfrac{1}{e} < 1$$. **Since, L is $$\dfrac{1}{e}$$ which is less than one, the series is convergent.** **Note:** In the above question instead of using ratio test we can directly find out by knowing the value of $$r$$ which is the common factor. Here in the above case $$r$$is $${e^{ - 1}} = \dfrac{1}{e}$$ . For a series to be convergent the range of $$r$$should be $$0 < |r| < 1$$. Since $$\dfrac{1}{e} < 1$$,the series is convergent.