# Does $a_n:=\left(1+\frac1{n+c}\right)^n$ converge to $e$, for $c\in\Bbb{N}$?

In my lecture, we defined $$e$$ as $$\lim_{n\to\infty}\left(1+\frac 1n\right)^n = \sum_{k=0}^\infty \frac 1{k!} = e$$

While playing around with those definitions, I stumbled onto $$a_n := \left(1 + \frac{1}{n +c}\right)^n$$ where $$c \in \mathbb{N}$$. From plotting it, it seems like it converges to $$e$$ as well, but slower if $$c$$ gets larger, but I can't find a proof for it.

Is my assumption true, and if so how would I go about proving it?

• quicker with your $c = - \frac{1}{2}$ Commented Jan 12, 2021 at 21:55
• The expansion I wrote confirms your observation and shows the non negligible role of $c$ in the story. Commented Jan 13, 2021 at 16:25

Consider:

$$\lim_{n\to\infty}\bigg(1 + \frac{1}{n+c}\bigg)^{n}=\lim_{n\to\infty}\frac{\displaystyle\bigg(1 + \frac{1}{n + c}\bigg)^{n+ c}}{\displaystyle\bigg(1 + \frac{1}{n+c}\bigg)^{c}}=\frac{\displaystyle\lim_{n\to\infty}\bigg(1 + \frac{1}{n + c}\bigg)^{n + c}}{ \displaystyle\lim_{n\to\infty}\bigg(1 + \frac{1}{n + c}\bigg)^{c}}$$

Because $$n + c\to\infty$$ as $$n\to\infty$$:

$$=\frac{e}{\displaystyle\lim_{n\to\infty}\bigg(1+\frac{1}{n + c}\bigg)^{c}} = \frac{e}{1^{c}}=\boxed{e}$$

For the simple scheme you are trying, the fastest possible convergence is $$c = - \frac{1}{2},$$ so that the sequence becomes $$b_n = \left( \frac{2n+1}{2n-1} \right)^n$$

This decreases to $$e.$$

$$e \approx 2.71828182845904523536028747135266249775724709369995$$

With $$n \geq 268$$ we $$b_n < 2.718285$$

With $$n \geq 1780$$ we $$b_n < 2.7182819$$

With $$n \geq 12023$$ we $$b_n < 2.71828183$$

267    2.718285006005104
268    2.71828498233662

1779    2.718281900038067
1780    2.718281899955931

12022    2.718281830026634
12023    2.718281829999742

19639    2.718281829072112
19640    2.718281828999756

40778    2.718281828526699
40779    2.718281828499953

45450    2.718281828512961
45451    2.718281828459387


Name $$c_n = \left( 1 + \frac{1}{n} \right)^n$$

Alright, from the Taylor series for log, we find $$b_n \approx e \; \cdot \; \; e^{\left( \frac{1}{12n^2} \right)}$$

while $$c_n \approx \frac{ e}{e^{\left( \frac{1}{2n} \right)} }$$ is a worse approximation

You can show this with a quick change of variables. Let $$m = n + c$$, which means that $$a_{n} = \left(1 + \frac{1}{m}\right)^{m-c} = \left(1 + \frac{1}{m}\right)^{m}\left(1 + \frac{1}{m}\right)^{-c}$$ Now, as $$m\to\infty$$ as $$n\to\infty$$, we have \begin{align} \lim_{n\to\infty}a_{n} &= \lim_{m\to\infty}\left(1 + \frac{1}{m}\right)^{m}\left(1 + \frac{1}{m}\right)^{-c}\\ &= \lim_{m\to\infty}\left(1 + \frac{1}{m}\right)^{m}\cdot\lim_{m\to\infty}\left(1 + \frac{1}{m}\right)^{-c}\\ &=e\cdot 1\\ &=e \end{align}

$$a_n=\left(1+\frac1{n+c}\right)^n\implies \log(a_n)=n \log\left(1+\frac1{n+c}\right)$$ $$\log\left(1+\frac1{n+c}\right)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\frac 1{(n+c)^k}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\frac 1{n^k}\frac 1{\left(1+\frac{c}{n}\right)^k }$$ $$\log\left(1+\frac1{n+c}\right)=\sum_{k=1}^\infty\frac{(-c)^k-(-c-1)^k}{k}\frac 1{n^k}$$ Computing the first terms $$\log\left(1+\frac1{n+c}\right)=\frac{1}{n}-\frac{2c+1}{2n^2}+\frac{3c^2+3c+1}{3n^3}+O\left(\frac{1}{n^4}\right)$$ $$\log(a_n)=1-\frac{2c+1}{2n}+\frac{3c^2+3c+1}{3n^2}+O\left(\frac{1}{n^3}\right)$$ $$a_n=e^{\log(a_n)}=e \left(1-\frac{2 c+1}{2 n}+\frac{36 c^2+36c +11}{24 n^2}\right)+O\left(\frac{1}{n^3}\right)$$

• good. I had wanted to make the point that a good choice for $c$ dramatically shrinks the error. Commented Jan 13, 2021 at 16:06
• @WillJagy. For sure ! For the expansion, we can see that $c$ is "important" for not too large values of $n$. The choise of $-\frac 12$ is particularly judicious. Cheers :-) Commented Jan 13, 2021 at 16:22