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$$\sum_{n=1}^{\infty} \left\{ e-\left(1+\frac{1}{n}\right)^n \right\}$$

Is this converge or diverge series .It is a series with positive terms ,but none of test of positive term series is seems to be working . How can we check ? Any hint?? Thanks in Advanced

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  • $\begingroup$ Rewrite $(1+1/n)^n = \exp(n \log(1 + 1/n))$, then extract an $\mathrm e$ out of the expression. $\endgroup$ – xbh Oct 21 '18 at 4:05
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    $\begingroup$ $a_n=e(1-e^{n\log(1+\frac{1}{n})-1})$ Then what can we do ?some sort of limit test ? $\endgroup$ – Eklavya Oct 21 '18 at 4:22
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\begin{align*} &e - \left(1 + \frac1n\right)^n \\ = \, &\left(\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \ldots\right) - \left(\binom{n}{0} + \frac{1}{n}\binom{n}{1} + \frac{1}{n^2}\binom{n}{2} + \ldots + \frac{1}{n^n}\binom{n}{n}\right) \\ = \, &\frac{1 - \left(1 - \frac{1}{n}\right)}{2!} + \frac{1 - \left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)}{3!} + \ldots + \frac{1 - \left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) \ldots \left(1 - \frac{n - 1}{n}\right)}{n!} \\ + \, &\frac{1}{(n + 1)!} + \frac{1}{(n + 2)!} + \ldots \\ \ge \, &\frac{1 - \left(1 - \frac{1}{n}\right)}{2!} = \frac{1}{2n}, \end{align*} hence the series diverges.

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Write $(1+ \frac{1}{n})^n=\exp(n \ln(1+\frac{1}{n})$, develop as $n \to \infty$ and find an equivalent of $e-(1+ \frac{1}{n})^n$.

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Note that $$ \newcommand\e{\mathrm e} \e - \left(1 +\frac 1n\right)^{n}= \e \left( 1 - \e^{n \log(1 + 1/n)-1}\right) = -\e \left(\exp \left(n \left(\frac 1n - \frac 1{2n^2} +o(n^{-2})\right)-1\right) -1\right) =-\e \left( \exp \left(- \frac 1{2n} + o(n^{-1})\right)-1\right)= \e \left(\frac 1{2n} + o(n^{-1})\right) \quad [n \to +\infty], $$ where we used the Maclaurin formulas. Now use the comparison test.

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As an alternative, by standard limits we have

$$\frac{e - \left(1 + \frac1n\right)^n}{\frac1n}=\frac{e - e^{\log\left(1 + \frac1n\right)^n}}{\frac1n}=e\cdot\frac{e^{\log\left[\left(1 + \frac1n\right)^n-1\right]}-1}{\log\left(1 + \frac1n\right)^n-1}\frac{1-\log\left(1 + \frac1n\right)^n}{\frac1n}\to\frac e 2$$

indeed

  • $t=\log\left[\left(1 + \frac1n\right)^n-1\right] \to 0$
  • $\frac{e^t-1}{t}\to 1$
  • $\frac{1-\log\left(1 + \frac1n\right)^n}{\frac1n}=\frac{\frac1n-\log\left(1 + \frac1n\right)}{\frac1{n^2}} \to \frac12$

therefore the given series diverges by limit comparison test with $\sum \frac1n$.

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Without power series. For $1\leq n\in \Bbb R:$

(I). We have $ \ln ((1+\frac {1}{n})^n)=$ $-n\ln (1-\frac {1}{n+1})=$ $n\int_{1-1/(n+1)}^1\frac {1}{y}dy<$ $< n\int_{1-1/(n+1)}^1 1\cdot dy=$ $\frac {n}{n+1}.$

So $(1+\frac {1}{n})^n<e^{n/(n+1)}.$ Hence we have

$\bullet \; e-(1+\frac {1}{n})^n>$ $e-e^{n/(n+1)}=$ $e\cdot \frac {1}{e^{1/(n+1)}}\cdot (e^{1/(n+1)}-1)>$ $>e\cdot \frac {1}{2}\cdot (e^{1/(n+1)}-1).$

(Because $\frac {1}{e^{1/(n+1)}}\geq$ $\frac {1}{e^{1/2}}>$ $\frac {1}{4^{1/2}}=$ $\frac {1}{2}.)$

(II). We have $ \frac {1}{n+1}=$ $\int _1^{1+1/(n+1)}1\cdot dy>$ $\int_1^{1+1/(n+1)}\frac {1}{y}dy=$ $\ln (1+\frac {1}{n+1}).$

So $e^{1/(n+1)}>$ $1+\frac {1}{n+1}$. Hence we have

$ \bullet \bullet \; e^{1/(n+1)}-1>\frac {1}{n+1}.$

(III). Therefore by $\bullet$ and $\bullet \bullet$ we have

$e-(1+\frac {1}{n})^n>$ $e\cdot\frac {1}{2}\cdot (e^{1/(n+1)}-1)>$ $e\cdot\frac {1}{2}\cdot \frac {1}{n+1}.$

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  • $\begingroup$ In (I) and (II) we are mainly just developing a special case (for $x\in \Bbb Z^+$) of $x>0\implies \frac {x}{1+x}<\ln (1+x)<x.$ $\endgroup$ – DanielWainfleet Oct 21 '18 at 7:54

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