Is $\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}$ konvergent or divergent?

Question

Is $$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}$$

konvergent or divergent?

I'm trying to use the squeeze theorem here but I can't find a suiting lower bound. For the upper bound I have that

$$\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}\leq n\cdot \frac{1}{n}=1.$$

Is $\frac{n}{\sqrt{n^2+n}}$ suitable for a lower bound? If so, please explain how. I can't see that this will always be less than or equal to my sum.

Answer 1

Hint: $$\sum_{k=1}^n\dfrac{1}{\sqrt{n^2+k}}\geq\sum_{k=1}^n\dfrac{1}{\sqrt{n^2+n}}=\dfrac{n}{\sqrt{n^2+n}}\to1$$ as $n\to\infty$.