Finding the mistake in the limit

Question

Please, help me with finding the mistake, where did i go wrong: $$L=\lim_{n \to \infty}(\frac{1}{\sqrt{n}\sqrt{n+1}}+\frac{1}{\sqrt{n}\sqrt{n+2}}+...+\frac{1}{\sqrt{n}\sqrt{n+n}}) $$

I tried the squeeze theorem, and I can see that $L \in [\frac{1}{2},1]$.

I found the solution, as $$L=\lim_{n\to\infty}\sum_{i=1}^n{\frac{1}{\sqrt{n}\sqrt{n+i}}}$$ $$L=\lim_{n\to\infty}\sum_{i=1}^n{\frac{1}{{n}\sqrt{1+\frac{i}{n}}}}$$

We can look at the limit as a way to calculate the area of a function of reals, so we have $$f(x)=\frac{1}{\sqrt{1+x}}$$ $$L=\int_{0}^1f(x)=...=2(\sqrt2-1)$$

Is there a way to do it without integration? Another way, a better way?

Second part of the question:

Trying another way of finding solution, I made a mistake that shows lack of fundamental understanding:

$$L=\lim_{n \to \infty}(\frac{1}{\sqrt{n}\sqrt{n+1}}+\frac{1}{\sqrt{n}\sqrt{n+2}}+...+\frac{1}{\sqrt{n}\sqrt{n+n}}) $$

Now we take $ln$ of both sides:

$$\ln{L}=\ln\lim_{n \to \infty}(\frac{1}{\sqrt{n}\sqrt{n+1}}+\frac{1}{\sqrt{n}\sqrt{n+2}}+...+\frac{1}{\sqrt{n}\sqrt{n+n}}) $$

$$\ln{L}=\lim_{n \to \infty}(\ln\frac{1}{\sqrt{n}\sqrt{n+1}}+\ln\frac{1}{\sqrt{n}\sqrt{n+2}}+...+\ln\frac{1}{\sqrt{n}\sqrt{n+n}}) $$

$$\ln{L}=-\lim_{n \to \infty}(\ln({\sqrt{n}\sqrt{n+1}})+\ln({\sqrt{n}\sqrt{n+2}})+...+\ln({\sqrt{n}\sqrt{n+n}})) $$

$$\ln{L}=-\lim_{n \to \infty}(\ln({\sqrt{n}\sqrt{n+1}})+\ln({\sqrt{n}\sqrt{n+2}})+...+\ln({\sqrt{n}\sqrt{n+n}})) $$

$$\ln{L}=-\lim_{n \to \infty}\ln{\sqrt{{(n^n)\prod_{i=1}^n(n+i)}}}$$

There are indeterminate forms here, but they are clearly diverging to infinity.

$$\ln{L}=-\infty$$

$$L=0$$ Where did I go wrong in the other procedure?

Answer 1

Since $$\sqrt{n+k+1}-\sqrt{n+k}=\frac1{\sqrt{n+k+1}+\sqrt{n+k}},$$ we have $$\frac12\frac1{\sqrt{n+k+1}}<\sqrt{n+k+1}-\sqrt{n+k}<\frac12\frac1{\sqrt{n+k}}$$ or $$2(\sqrt{n+k+1}-\sqrt{n+k})<\frac1{\sqrt{n+k}}<2(\sqrt{n+k}-\sqrt{n+k-1}).$$ This means $$2(\sqrt{2n+1}-\sqrt{n+1})<\sum^n_{k=1}\frac1{\sqrt{n+k}}<2(\sqrt{2n}-\sqrt{n}),$$ i.e. $$2\left(\sqrt{2+1/n}-\sqrt{1+1/n}\right)<\sum^n_{k=1}\frac1{\sqrt{n}}\frac1{\sqrt{n+k}}<2(\sqrt{2}-1).$$

Answer 2

This is just a comment added for your curiosity.

$$S_n=\sum_{i=1}^n{\frac{1}{\sqrt{n}\sqrt{n+i}}}=\frac{1}{\sqrt{n}}\sum_{i=1}^n{\frac{1}{\sqrt{n+i}}}=\frac{1}{\sqrt{n}}\left(\zeta \left(\frac{1}{2},n+1\right)-\zeta \left(\frac{1}{2},2 n+1\right) \right)$$ where appears the Hurwitz zeta function (don't worry : you will learn about it).

Using the asymptotics $$\zeta \left(\frac{1}{2},q+1\right)=-2 \sqrt{q}-\frac{1}{2 \sqrt{q}}+\frac{1}{24 q^{3/2}}-\frac{1}{384 q^{7/2}}+O\left(\frac{1}{q^{9/2}}\right)$$ and continuing with Taylor series we then find $$S_n=2 \left(\sqrt{2}-1\right)-\frac{2-\sqrt{2}}{4\, n}+\frac{4-\sqrt{2}}{96\, n^2}-\frac{16-\sqrt{2}}{6144\, n^4}+O\left(\frac{1}{n^6}\right)$$ which is extremely good even for small values of $n$ as shown below $$\left( \begin{array}{ccc} n & \text{exact} & \text{approximation} \\ 1 & 0.7071067812 & 0.7065418021 \\ 2 & 0.7618016811 & 0.7617892646 \\ 3 & 0.7825762847 & 0.7825750882 \\ 4 & 0.7934898748 & 0.7934896537 \\ 5 & 0.8002114745 & 0.8002114155 \\ 6 & 0.8047657468 & 0.8047657268 \\ 7 & 0.8080548993 & 0.8080548913 \\ 8 & 0.8105415863 & 0.8105415827 \\ 9 & 0.8124874535 & 0.8124874517 \\ 10 & 0.8140515801 & 0.8140515792 \end{array} \right)$$