# Convergence of the series $\sum_{n=1}^{\infty}(-1)^n(1+\frac{1}{n})^{n+1}$

Does $$\sum_{n=1}^{\infty}(-1)^n(1+\frac{1}{n})^{n+1}$$ converge.

According to Wolfram Alpha, the sum converges to a value $$-2.239...$$

Problem:: But isn't it true that the limit of the absolute value of the terms is $$e$$ and not zero? I thought this means it cannot converge.

Thanks.

• You are right. The series does not converge. Nov 18, 2019 at 0:03
• But how can I go against all know wolfram alpha?wolframalpha.com/input/… Nov 18, 2019 at 0:05
• For what it's worth, WolframAlpha (i) is not the truth, and makes mistakes/interprets the input in sometimes unexpected ways; (ii) in this particular case, does sometimes return "sum does not converge" in my case. (roughly every other time I click on the link.) Nov 18, 2019 at 0:31
• Nov 18, 2019 at 0:36

Let $$(a_n)=\sum_{m=1}^n (-1)^m\left(1+\frac{1}{m}\right)^{m+1}$$
As pointed out by Kabo Murphy, the series does diverge, that is $$(a_n)$$ diverges. The reason Wolfram alpha said $$(a_n)\to -2.239...$$ is because $$(a_n)$$ has two convergent subsequences: $$(a_{2n})\to -0.880...$$ and $$(a_{2n+1})\to -3.599...$$. Taking the average of the two, $$(a_n)\to-2.239...$$ which is how Wolfram Alpha probably got the solution.
• May I ask why $(a_{2n})$ and $(a_{2n + 1})$ converge? Nov 18, 2019 at 1:24
• @Azlif $$(a_{2n})=\sum_{m=1}^n \left(\left(1+\frac{1}{2m}\right)^{2m+1}-\left(1+\frac{1}{2m-1}\right)^{2m}\right)$$ $$(a_{2n+1})=-\left(1+\frac{1}{2n+1}\right)^{\left(2n+1\right)}+(a_{2n})$$ You can use the ratio test to show that $(a_{2n})$ converges (then $(a_{2n+1})$ converges as well). $(a_{2n})$ was obtained by writing out the sum term by term and grouping every two terms together. Nov 18, 2019 at 14:45