Find the series:

$\sum _{n=1}^{\mathrm{\infty}}({(1+\frac{1}{n})}^{n}-e)$

Jamarion Kerr
2022-01-22
Answered

Find the series:

$\sum _{n=1}^{\mathrm{\infty}}({(1+\frac{1}{n})}^{n}-e)$

You can still ask an expert for help

immablondevl

Answered 2022-01-23
Author has **11** answers

Take the negative of that series to make the terms positive and limit compare it with $\frac{1}{n}$

$\underset{n\to \mathrm{\infty}}{lim}\frac{e-{(1+\frac{1}{n})}^{n}}{\frac{1}{n}}\to \frac{0}{0}$

$\Rightarrow \underset{n\to \mathrm{\infty}}{lim}\frac{-{(1+\frac{1}{n})}^{n}(\mathrm{log}(1+\frac{1}{n})-\frac{1}{n+1})}{-\frac{1}{{n}^{2}}}$

which we get from LHopital

which we get from LHopital

gekraamdbk

Answered 2022-01-24
Author has **13** answers

We first observe, that

${(1+\frac{1}{n})}^{n}-e={e}^{n\mathrm{ln}(1+\frac{1}{n})}-e$

Elementary calculus arguments can show that the function$x\mathrm{ln}(1+\frac{1}{x})$ is monotone increasing on $(0,\mathrm{\infty})$ and that its range has 1 as a supremum and has 0 as an infimum.

We now invoke the Mean Value Theorem to conclude that for every$n\in \mathbb{N}$ there is a real $\zeta}_{n$ between 1 and $n\mathrm{ln}(1+\frac{1}{n})$ such that

$e-{(1+\frac{1}{n})}^{n}={e}^{{\zeta}_{n}}(1-n\mathrm{ln}(1+\frac{1}{n}))$

From what we know about$x\mathrm{ln}(1+\frac{1}{x})$ , we can conclude that

${e}^{0}(1-n\mathrm{ln}(1+\frac{1}{n}))\le e-{(1+\frac{1}{n})}^{n}\le {e}^{1}(1-n\mathrm{ln}(1+\frac{1}{n}))$

Now we use the inequalities

$\frac{1}{2(1+x)}\le 1-x\mathrm{ln}(1+\frac{1}{x})\le \frac{1}{x+1}$

which is true for all positive x to conclude

$\frac{1}{2(1+n)}\le e-{(1+\frac{1}{n})}^{n}\le \frac{e}{1+n}$

which will carry the rest of the argument.

The real crux to this argument are the inequalities

$\frac{1}{2(1+x)}\le 1-x\mathrm{ln}(1+\frac{1}{x})\le \frac{1}{x+1}$

The right-hand inequality is equivalent to the more familiar inequality

$\mathrm{ln}(1+x)\le x$

which I can derive if needed.

The left-hand inequality is equivalent to the slightly stronger inequality

$\mathrm{ln}(1+x)\le x-\frac{{x}^{2}}{2(1+x)}$

which is however established in very much the same way as the weaker inequality.

Elementary calculus arguments can show that the function

We now invoke the Mean Value Theorem to conclude that for every

From what we know about

Now we use the inequalities

which is true for all positive x to conclude

which will carry the rest of the argument.

The real crux to this argument are the inequalities

The right-hand inequality is equivalent to the more familiar inequality

which I can derive if needed.

The left-hand inequality is equivalent to the slightly stronger inequality

which is however established in very much the same way as the weaker inequality.

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