Find $\lim\limits_{n \to \infty} \left ( n - \sum\limits_{k = 1} ^ n e ^{\frac{k}{n^2}} \right)$.

Question

I have to find the limit:

$$\lim\limits_{n \to \infty} \bigg ( n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \bigg)$$

This is what I managed to do:

$$ e^{\frac{1}{n^2}} + e^{\frac{1}{n^2}} + ... + e^{\frac{1}{n^2}} \le e^{\frac{1}{n^2}} + e^{\frac{2}{n^2}} + ...e^{\frac{n}{n^2}} \le e^{\frac{n}{n^2}} + e^{\frac{n}{n^2}} + ... e^{\frac{n}{n^2}}$$

$$ n e^{\frac{1}{n^2}} \le \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le ne^{\frac{1}{n}}$$

$$ -n e^{\frac{1}{n}} \le - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le - n e^{\frac{1}{n^2}}$$

$$ n - n e^{\frac{1}{n}} \le n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le n - n e^{\frac{1}{n^2}}$$

Here I found that the limit of the left-hand side is equal to $-1$, while the limit of the right-hand side is $0$. So I got that:

$$-1 \le n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le 0$$

And I cannot draw a conclusion about the exact limit. What should I do?

Answer 1

Hint: $$a+a^2+...+a^n=\frac{a(a^n-1)}{a-1}$$

Answer 2

Using $e^x=1+x+O(x^2)$ along with $\sum_{k=1}^nk=\frac{n(n+1)}{2}$ and $\sum_{k=1}^n k^2=O\left(n^3\right)$ , we assert that

$$\begin{align} \sum_{k=1}^ne^{k/n^2}&=\sum_{k=1}^n\left(1+\frac{k}{n^2}\right)+O\left(\frac{1}{n}\right)\\\\ &=n+\frac{n(n+1)}{2n^2}+O\left(\frac1n\right)\\\\ &=n+\frac12+O\left(\frac1n\right) \end{align}$$

Hence, we see that

$$\lim_{n\to\infty}\left(n-\sum_{k=1}^ne^{k/n^2}\right)=-\frac12$$

Answer 3

As @Andrei already answered, you face a geometric sum, that is to say that $$\sum_{k = 1} ^ n e ^{\frac{k}{n^2}}=\sum_{k = 1} ^ n \left(e ^{\frac{1}{n^2}}\right)^k=\frac{e^{\frac{1}{n^2}} \left(e^{\frac{1}{n}}-1\right)}{e^{\frac{1}{n^2}}-1}$$ Now, expanding the exponentials as Taylor series, multiplying for the numerator and then long division, you should get $$\sum_{k = 1} ^ n e ^{\frac{k}{n^2}}=n+\frac{1}{2}+\frac{2}{3 n}+O\left(\frac{1}{n^2}\right)$$ $$ n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}}=-\frac{1}{2}-\frac{2}{3 n}+O\left(\frac{1}{n^2}\right) $$ which shows the limit and also how it is approached.

Edit

In comments, @marty cohen made a good point.

If we consider $$n-\sum_{k = 1} ^ {n^a} e ^{\frac{k}{n^2}}=n-\frac{e^{\frac{1}{n^2}} \left(e^{n^{a-2}}-1\right)}{e^{\frac{1}{n^2}}-1}$$ it will not converge for any $a \neq 1$ since it will be $(n-n^a+\cdots)$