# $\sum_{n=1}^{\infty} \left\{ e-(1+\frac{1}{n})^n \right\}$. is this converge or diverge

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

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

• Rewrite $(1+1/n)^n = \exp(n \log(1 + 1/n))$, then extract an $\mathrm e$ out of the expression. – xbh Oct 21 '18 at 4:05
• $a_n=e(1-e^{n\log(1+\frac{1}{n})-1})$ Then what can we do ?some sort of limit test ? – Eklavya Oct 21 '18 at 4:22

\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.

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

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.

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 $$\sum \frac1n$$.

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 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

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

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