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Because $(1+(1/n))^n$ converges to $e$, I was thinking of comparing the sum to the series $p^{\sqrt{\log(n)}}$, but this series apparently diverges according to wolfram so I'm at a loss. Can someone give me a clue as to where to go from here?

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  • $\begingroup$ what's the background of this question? is there any reason you believe it should converge? $\endgroup$ – Dando18 Sep 15 '17 at 21:35
  • $\begingroup$ I'm simply trying to show it either converges or diverges. I've only been comparing it to series' that converge because I can't think of any to compare it to that might to divergence of the series. $\endgroup$ – qqmuffin Sep 15 '17 at 21:38
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    $\begingroup$ $(1+1/n)^n - e < 0$ while $\sqrt{\log n}$ is irrational, so the terms don't even make sense (except possibly as complex numbers). $\endgroup$ – Daniel Schepler Sep 15 '17 at 21:41
  • $\begingroup$ Assuming you meant $e-(1+1/n)^n$ for the reasons Daniel Scheplet points out, use the fact that your exponent is eventually larger than $2$ (say), so you only need something like $e-(1+1/n)^n=O(1/n)$. $\endgroup$ – Jason Sep 15 '17 at 21:48
  • $\begingroup$ This inequality may be useful: $1 - \frac1x \leq \log x \leq x-1 \quad\text{for all } x > 0$}. $\endgroup$ – MathOverview Sep 15 '17 at 21:49
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I will assume that the absolute value of $\left(1+\tfrac{1}{n}\right)^n-e$ is taken since otherwise the terms don't exist, as mentioned in the comment.

The difference between $e$ and $\left(1+\tfrac{1}{n}\right)^n$ is of order $\tfrac{1}{n}$ (expand the binomial to the second order term to see that; this is mentioned in the comments, too). So for large $n$ the terms are of order $n^{-\sqrt{\log(n)}}$. For $n$ large enough, they are bound from above by, say, $n^{-2}$; hence, the series converges.

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Hint: Prove that

$$\lim_{n\to\infty} \frac{e-\left(1+\frac{1}{n}\right)^n}{\frac{e}{2n}} = 1.$$

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Note that \begin{align} \left( 1+\frac{1}{n} \right)^{n} =& \frac{n!}{0!(n-0)!}\frac{1}{n^0} + \frac{n!}{1!(n-1)!}\frac{1}{n} + \cdots + \frac{n!}{n!(n-n)!}\frac{1}{n^n} \\ =& \frac{1}{0!} + \frac{1}{1!}\left( 1-\frac{1}{n}\right) + \frac{1}{2!}\left( 1-\frac{1}{n}\right)\left( 1-\frac{2}{n}\right) + \cdots + \frac{1}{n!}\left( 1-\frac{1}{n}\right)\cdot \cdots\cdot \left( 1-\frac{n-1}{n}\right) \\ =& \sum_{k=0}^{n} \frac{1}{k!} \prod_{i=0}^{k} \left( 1-\frac{i}{n} \right) \end{align} implies $$ \left| \; e-\left(1+\frac{1}{n} \right)^n \;\right| = \sum_{k>n} \frac{1}{k!} \prod_{i=1}^{k} \left( 1-\frac{i}{n} \right) \leq \sum_{k>n} \frac{1}{k!} =\frac{1}{n^2} \sum_{k>n} \frac{n^2}{k!} $$ Here $\prod_{i=0}^{0} \left( 1-\frac{k}{n} \right)=1$. Since $n$ is constant with respect to $k$, $ \sum_{k>n} \frac{n^2}{k!} $ is limited say by a constant $L$. Then we have $$ \left| \; e-\left(1+\frac{1}{n} \right)^n \;\right| <\frac{1}{n^2}\cdot L. $$ Recall that $1 - \frac1x \leq \log x \leq x-1$ for all $ x > 0$. Then \begin{align} \sum_{n=2}^{\infty} \left|e - \left( 1+\frac{1}{n} \right)^n \right|^{\sqrt{\log(n)}} &\leq \sum_{n=2}^{\infty} \left(\frac{L}{n^2}\right)^{\sqrt{\log(n)}} \\ &\leq \sum_{n=2}^{\infty} \left(\frac{L}{n^2}\right)^{\sqrt{1-1/n}} \\ &\leq \sum_{n=2}^{\infty} \left(\frac{L}{n^2}\right)^{1/4} \\ &= \sum_{n=2}^{\infty} \left(\frac{L}{n^{3/4}}\right) \end{align}

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  • $\begingroup$ Can you explain how the last two steps connect? I was under the impression that since 0<ϵ<1, the smaller exponent would result in a larger value. $\endgroup$ – qqmuffin Sep 16 '17 at 0:03
  • $\begingroup$ @qqmuffin Now I belive that may answer is correct. $\endgroup$ – MathOverview Sep 16 '17 at 14:06
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Do you mean $\enspace\displaystyle\sum_{n=1}^\infty \left(e-\left(1+\frac{1}{n}\right)^n \right)^{\sqrt{\ln n}}\enspace$ ?

If Yes : See $\enspace\displaystyle\sum_{n=1}^\infty \left(e-\left(1+\frac{1}{n}\right)^n \right)^2\enspace$ (convergent) in

Closed form for $\sum_{n=1}^\infty \left(e-\left(1+\frac{1}{n}\right)^n \right)^2$? .

We have $\enspace\displaystyle 0<\sum_{n=55}^\infty \left(e-\left(1+\frac{1}{n}\right)^n \right)^{\sqrt{\ln n}}<\sum_{n=55}^\infty \left(e-\left(1+\frac{1}{n}\right)^n \right)^2\enspace$ and $\left(\left(e-\left(1+\frac{1}{n}\right)^n \right)^{\sqrt{\ln n}}\right)_n$ is a null sequence therefore we have convergence.

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In this answer it is shown that $\left(1+\frac1n\right)^n$ is increasing and $\left(1+\frac1n\right)^{n+1}$ is decreasing. Therefore, $$ \left(1+\frac1n\right)^n\le e\le\left(1+\frac1{n-1}\right)^n\tag{1} $$ Using the formula for $\frac{a^n-b^n}{a-b}$, we get $$ \begin{align} \frac{\left(1+\frac1{n-1}\right)^n-\left(1+\frac1n\right)^n}{\frac1{n-1}-\frac1n} &=\sum_{k=1}^n\left(1+\frac1{n-1}\right)^{n-k}\left(1+\frac1n\right)^{k-1}\\[6pt] &\le n\left(1+\frac1{n-1}\right)^{n-1}\\[12pt] &\le ne\tag{2} \end{align} $$ Therefore, $$ \begin{align} e-\left(1+\frac1n\right)^n &\le\left(1+\frac1{n-1}\right)^n-\left(1+\frac1n\right)^n\\ &\le\frac{e}{n-1}\tag{3} \end{align} $$


$$ \begin{align} \sum_{n=2}^\infty\left(e-\left(1+\frac1n\right)^n\right)^{\sqrt{\log(n)}} &\le\left(e-\frac94\right)^{\sqrt{\log(2)}}+\sum_{n=3}^\infty\left(\frac{e}{n-1}\right)^{\sqrt{\log(n)}}\\ &\le\left(e-\frac94\right)^{\sqrt{\log(2)}}+\sum_{n=3}^\infty\left(\frac{e}{n-1}\right)^{\sqrt{\log(3)}}\tag{4} \end{align} $$ where the sum converges because $\sqrt{\log(3)}\gt1$.

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