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Is the series $$\sum_{n=1}^\infty\left(1-\frac1{\sqrt n}\right)^n$$

convergent?

Schwale
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  • $\left(1-\frac1{\sqrt n}\right)^n \sim e^{-\sqrt{n}}$, so the sequence converges. – r9m Dec 03 '15 at 21:21
  • Why is that so obvious? – Schwale Dec 03 '15 at 21:23
  • I guess my answer makes no sense, but nobody gives a reason...Ridiculous that downvotes are given without revealing who it is... – imranfat Dec 03 '15 at 21:26
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    @imranfat I had downvoted: from what you wrote, it was not clear to me how the hint helped the OP (after the step you gave, it remained to deal with the exponentiation). – Clement C. Dec 03 '15 at 21:31
  • I appreciate the honesty. I deleted my answer, because after furtering my computations, the Root test gave me 1, so that was of no use after all... – imranfat Dec 03 '15 at 21:33
  • (Also, after you delete your answer, impossible to comment on it.) @r9m: there is a constant factor missing in your equivalent ($e^{-1/2}$), even though this does not change anything important. – Clement C. Dec 03 '15 at 21:34
  • @ClementC. Yes of course, I guess I should have written $O$ (big-oh). – r9m Dec 03 '15 at 21:37

3 Answers3

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Rewrite $$ a_n = \left(1-\frac{1}{\sqrt n}\right)^n = e^{n\ln\left(1-\frac1{\sqrt n}\right) } $$ and using the Taylor series (to order two) of $\ln(1-x)$ around $0$, $\ln(1-x) = -x-\frac{x^2}{2} + o(x^2)$, you obtain $$ a_n = e^{-n\ln\left(\frac1{\sqrt n} + \frac1{2n} + o(\frac{1}{n})\right) } = e^{-\sqrt n - \frac{1}{2} + o(1) } \operatorname*{\sim}_{n\to\infty} e^{- \frac{1}{2}}e^{-\sqrt n}. $$ To conclude, theorems of comparison (between non-negative series) ensure that the two series $\sum_n a_n$ and $\sum_n e^{-\sqrt n}$ have same nature.

Clement C.
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Outline: Since $\left(1-\frac{1}{\sqrt{n}}\right)^{\sqrt{n}}$ has limit $\frac{1}{e}$, after a while (indeed immediately) it is $\le \frac{1}{2}$.

So the $n$-th term of our sequence is $\le \frac{1}{2^{\sqrt{n}}}$.

After a while, this is less than $\frac{1}{n^2}$, since $\ln(2^{\sqrt{n}})=\sqrt{n}(\ln 2)$ and $\ln(n^2)=2\ln n$, and after a while $\sqrt{n}(\ln 2)$ is (much) bigger than $2\ln n$.

But $\sum \frac{1}{n^2}$ converges, and therefore by comparison so does our series.

André Nicolas
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Hint: the upper bound $$ \left(1-\frac1{\sqrt n}\right)\le e^{-\frac1{\sqrt n}} $$ follows directly from the inequality $1+x\le e^x$.

Community
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grand_chat
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