# How do I evaluate $\lim_{n\to\infty} \,\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}$? [closed]

I came across the following problem recently in a problem sheet aimed at high school students:

Evaluate $$\lim_{n\to\infty} \,\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}.$$

I tried to rewrite the inner sum as a Riemann sum hoping that the limit would become a definite integral, but no gain because of the extra $$1/n$$'s.

We can rewrite the sum as

$$S = \lim_{n\to\infty}\sum_{k=1}^n \left(\frac{k}{n^2}\right)^{\frac{k}{n^2}}\cdot\frac{k}{n}\cdot\frac{1}{n}$$

We also have that for $$1\leq k \leq n$$

$$\left(\frac{1}{n}\right)^{\frac{1}{n}} \leq \left(\frac{k}{n^2}\right)^{\frac{k}{n^2}} \leq \left(\frac{1}{n^2}\right)^{\frac{1}{n^2}}$$

for $$n > e$$. Thus we can sandwich the original limit

$$\lim_{n\to\infty} \left(\frac{1}{n}\right)^{\frac{1}{n}} \cdot \sum_{k=1}^n \frac{k}{n}\cdot\frac{1}{n} \leq S \leq \lim_{n\to\infty} \left(\frac{1}{n^2}\right)^{\frac{1}{n^2}} \cdot \sum_{k=1}^n \frac{k}{n}\cdot\frac{1}{n}$$

which means that

$$S = \lim_{n\to\infty} \sum_{k=1}^n \frac{k}{n}\cdot\frac{1}{n} = \int_0^1x\:dx = \frac{1}{2}$$

by squeeze theorem.

• You don't want to take the limit in the last inequality.
– zhw.
Commented Jun 26, 2020 at 17:51
• @zhw could you explain why? Commented Jun 26, 2020 at 19:46
• When you take the limit you get equality, not strict inequality.
– zhw.
Commented Jun 26, 2020 at 19:47
• The condition for squeeze theorem is $\leq$ rather than <.
– dan
Commented Jun 29, 2020 at 4:27

For those who didn't understand why, for $$n>e$$, $$\left(\frac{1}{n}\right)^{\frac{1}{n}} < \left(\frac{k}{n^2}\right)^{\frac{k}{n^2}} < \left(\frac{1}{n^2}\right)^{\frac{1}{n^2}}$$, the graph of $$f(x)=x^x$$ starts increasing after $$x=1/e$$ for $$f^\prime(x)=(1+\ln x)e^{x\ln x}=(1+\ln x)x^x$$.