Evaluating $\lim_{n\to \infty}\left(\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\dots +\frac{1}{\sqrt{n^2+2n}}\right)$

Question

Question:

Evaluate the following limit: $$\lim_{n\to \infty}\left(\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\dots +\frac{1}{\sqrt{n^2+2n}}\right)$$

My Approach:

The first step I did was to split the limits using the following property:

$$\lim_{x \to a} \left(f(x)+g(x) \right) = \lim_{x \to a}f(x)+\lim_{x \to a}g(x)$$

Like this:

$$\lim_{n\to \infty}\left(\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\dots +\frac{1}{\sqrt{n^2+2n}}\right) = $$ $$\lim_{n\to \infty}\left(\frac{1}{\sqrt{n^2}}\right)+\lim_{n\to \infty}\left(\frac{1}{\sqrt{n^2+1}}\right)+\lim_{n\to \infty}\left(\frac{1}{\sqrt{n^2+2}}\right)+\dots+\lim_{n\to \infty}\left(\frac{1}{\sqrt{n^2+2n}}\right)$$

We know that, $$\lim_{n\to \infty}\left(\frac{1}{n} \right)=0$$

Applying the same concept to all the individual limits obtained, the answer must be $0$. But the answer in my text book is given to be $2$. Did I go wrong some where, or, is the answer in the textbook incorrect?

I don't think I am wrong, because even the largest term (the one which has the comparatively smaller denominator), i.e, the first term in the summation, itself is tending towards zero. So rest of the terms must be much closer to zero. Closer to zero means very close to zero and hence each term must be equal to zero as indicated by the property, and so the entire limit must tend towards zero. But the answer says the value of the limit is $2$

Please explain how to solve* this problem and, where and why did I go wrong.

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*I cant think of any other method of solving this problem other than the one I specified above.

Answer 1

You can write the limit as:

$$\lim\limits_{n\rightarrow +\infty }\sum\limits_{i=0}^{2n} \frac{1}{\sqrt{n^2 +i }}$$

The problem with your argument is that the numbers you are adding is also going to infinity.

To get the result, notice that

$$\sum\limits_{i=0}^{2n} \frac{1}{\sqrt{n^2 +i }} < \sum\limits_{i=0}^{2n} \frac{1}{\sqrt{n^2 }} = 2$$

And that

$$\sum\limits_{i=0}^{2N} \frac{1}{\sqrt{n^2 +i }} > \sum\limits_{i=0}^{2N} \frac{1}{\sqrt{n^2 +2n}} = \frac{2n}{\sqrt{n^2 +2n}} = \frac{1}{\sqrt{\frac{1}{4} +\frac{1}{2n}}}$$

The limit of the last term is $2$. Then, using the "Sandwich Theorem", $$\lim\limits_{n\rightarrow +\infty }\sum\limits_{i=0}^{2n} \frac{1}{\sqrt{n^2 +i }} =2$$

Answer 2

$$\lim_{n\rightarrow \infty} \sum_{k=0}^{2n} \frac{1}{n} \frac{1}{\sqrt{1+k/n^2}}= \int_{0}^{2} dx =2.$$ Here $\frac{k}{n^2}-\rightarrow 0.$