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.


  • $\begingroup$ You are right. The series does not converge. $\endgroup$ Nov 18, 2019 at 0:03
  • $\begingroup$ But how can I go against all know wolfram alpha?wolframalpha.com/input/… $\endgroup$ Nov 18, 2019 at 0:05
  • 3
    $\begingroup$ 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.) $\endgroup$
    – Clement C.
    Nov 18, 2019 at 0:31
  • $\begingroup$ Also: wolframalpha.com/input/… $\endgroup$
    – Clement C.
    Nov 18, 2019 at 0:36

1 Answer 1


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.

  • $\begingroup$ May I ask why $(a_{2n})$ and $(a_{2n + 1})$ converge? $\endgroup$
    – Azlif
    Nov 18, 2019 at 1:24
  • 1
    $\begingroup$ @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. $\endgroup$
    – Tom Himler
    Nov 18, 2019 at 14:45

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