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What test do i use to show this series converges or diverges?

$$\sum_{r=1}^{\infty}\frac{1}{(1+\frac{1}{r})^{r}}$$

I know that $(1+\frac{1}{r})^{r} \rightarrow e$ so does this function converge to $\frac{1}{e}$? and is it sufficient just to say that?

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  • $\begingroup$ Why is there an equal to sign in between sigma and the function ??? $\endgroup$ – Murtuza Vadharia Apr 22 '16 at 17:21
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Yes, looking at the limit you gave is sufficient: the series diverges. It is a basic (yet fundamental) result that if the series $\sum_n a_n$ converges, then it must be the case that $a_n \xrightarrow[n\to\infty]{}0$.

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You have determined that the sequence $$a_r=\frac{1}{(1+\frac{1}{r})^r}$$ converges to $\frac{1}{e}$.

This is a sufficient condition to conclude that the series $$\sum_{r=1}^{\infty}{a_r}=\sum_{r=1}^{\infty}{\frac{1}{(1+\frac{1}{r})^r}}$$ diverges.

In order for a series to converge, it is necessary that the terms within the series converge to $0$. This should make sense if you consider that, in the above case, as r approaches infinity, $\frac{1}{e}$ must be continually added to the total (and it will therefore keep growing). In order for the series to converge, eventually, you must be adding $0$'s to the total.

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