# 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:

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.

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

• Your calculations are not right, since we have in your case infinity many terms. – Dr. Sonnhard Graubner Sep 25 '19 at 17:10
• Usually textbook is right. Limit of sum is not always equal to sum of limits (try to replace each term with $1/n$, for example). – Vasya Sep 25 '19 at 17:10
• The problem you've run into is the classic Riemann sum fallacy: you're adding an ever increasing number of terms, each individually shrinking to $0$. However, in the limit, the Riemann sums don't necessarily tend to $0$, they tend to a definite integral. While you can split the limit over a sum of two sequences, or some fixed, finite number of sequences, the sum is expanding, which the algebra of limits does not cover. – Theo Bendit Sep 25 '19 at 17:10
• As a more concrete example, take the sum of $\frac{1}{n}$ $n$ times. Each of the $n$ terms approaches $0$, but the sum is constantly $1$, and hence $1$ in the limit. – Theo Bendit Sep 25 '19 at 17:11
• @TheoBendit, Thank you for your reply. I understood my mistake. – Guru Vishnu Sep 25 '19 at 17:13

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

• Thank you for your excellent answer. It would be great if you could specify that you have used the 'Sandwich Theorem'. – Guru Vishnu Sep 25 '19 at 17:22
• Ok, edited. Thanks! – Jorge Catepillan Sep 25 '19 at 18:25

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