Jamarion Kerr

2022-01-22

Find the series:

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

immablondevl

Beginner2022-01-23Added 11 answers

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

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

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

which we get from LHopital

which we get from LHopital

gekraamdbk

Beginner2022-01-24Added 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.

What is the derivative of the work function?

How to use implicit differentiation to find $\frac{dy}{dx}$ given $3{x}^{2}+3{y}^{2}=2$?

How to differentiate $y=\mathrm{log}{x}^{2}$?

The solution of a differential equation y′′+3y′+2y=0 is of the form

A) ${c}_{1}{e}^{x}+{c}_{2}{e}^{2x}$

B) ${c}_{1}{e}^{-x}+{c}_{2}{e}^{3x}$

C) ${c}_{1}{e}^{-x}+{c}_{2}{e}^{-2x}$

D) ${c}_{1}{e}^{-2x}+{c}_{2}{2}^{-x}$How to find instantaneous velocity from a position vs. time graph?

How to implicitly differentiate $\sqrt{xy}=x-2y$?

What is 2xy differentiated implicitly?

How to find the sum of the infinite geometric series given $1+\frac{2}{3}+\frac{4}{9}+...$?

Look at this series: 1.5, 2.3, 3.1, 3.9, ... What number should come next?

A. 4.2

B. 4.4

C. 4.7

D. 5.1What is the derivative of $\frac{x+1}{y}$?

How to find the sum of the infinite geometric series 0.9 + 0.09 + 0.009 +…?

How to find the volume of a cone using an integral?

What is the surface area of the solid created by revolving $f\left(x\right)={e}^{2-x},x\in [1,2]$ around the x axis?

How to differentiate ${x}^{\frac{2}{3}}+{y}^{\frac{2}{3}}=4$?

The differential coefficient of $\mathrm{sec}\left({\mathrm{tan}}^{-1}\left(x\right)\right)$.