Jamarion Kerr

Answered question

2022-01-22

Find the series:
$\sum _{n=1}^{\mathrm{\infty }}\left({\left(1+\frac{1}{n}\right)}^{n}-e\right)$

Answer & Explanation

immablondevl

Beginner2022-01-23Added 11 answers

Take the negative of that series to make the terms positive and limit compare it with $\frac{1}{n}$
$\underset{n\to \mathrm{\infty }}{lim}\frac{e-{\left(1+\frac{1}{n}\right)}^{n}}{\frac{1}{n}}\to \frac{0}{0}$
$⇒\underset{n\to \mathrm{\infty }}{lim}\frac{-{\left(1+\frac{1}{n}\right)}^{n}\left(\mathrm{log}\left(1+\frac{1}{n}\right)-\frac{1}{n+1}\right)}{-\frac{1}{{n}^{2}}}$
which we get from LHopital

gekraamdbk

Beginner2022-01-24Added 13 answers

We first observe, that
${\left(1+\frac{1}{n}\right)}^{n}-e={e}^{n\mathrm{ln}\left(1+\frac{1}{n}\right)}-e$
Elementary calculus arguments can show that the function $x\mathrm{ln}\left(1+\frac{1}{x}\right)$ is monotone increasing on $\left(0,\mathrm{\infty }\right)$ 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 $n\mathrm{ln}\left(1+\frac{1}{n}\right)$ such that
$e-{\left(1+\frac{1}{n}\right)}^{n}={e}^{{\zeta }_{n}}\left(1-n\mathrm{ln}\left(1+\frac{1}{n}\right)\right)$
From what we know about $x\mathrm{ln}\left(1+\frac{1}{x}\right)$, we can conclude that
${e}^{0}\left(1-n\mathrm{ln}\left(1+\frac{1}{n}\right)\right)\le e-{\left(1+\frac{1}{n}\right)}^{n}\le {e}^{1}\left(1-n\mathrm{ln}\left(1+\frac{1}{n}\right)\right)$
Now we use the inequalities
$\frac{1}{2\left(1+x\right)}\le 1-x\mathrm{ln}\left(1+\frac{1}{x}\right)\le \frac{1}{x+1}$
which is true for all positive x to conclude
$\frac{1}{2\left(1+n\right)}\le e-{\left(1+\frac{1}{n}\right)}^{n}\le \frac{e}{1+n}$
which will carry the rest of the argument.
The real crux to this argument are the inequalities
$\frac{1}{2\left(1+x\right)}\le 1-x\mathrm{ln}\left(1+\frac{1}{x}\right)\le \frac{1}{x+1}$
The right-hand inequality is equivalent to the more familiar inequality
$\mathrm{ln}\left(1+x\right)\le x$
which I can derive if needed.
The left-hand inequality is equivalent to the slightly stronger inequality
$\mathrm{ln}\left(1+x\right)\le x-\frac{{x}^{2}}{2\left(1+x\right)}$
which is however established in very much the same way as the weaker inequality.

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