# How to solve this limit $\lim_{n\to\infty} ((\frac{1}{\sqrt{n^2+1}}) + \cdots + (\frac{1}{\sqrt{n^2+n}}))$

How do I solve this limit $$\lim_{n\to\infty} \left(\frac{1}{\sqrt{n^2+1}} + \cdots + \frac{1}{\sqrt{n^2+n}}\right)\text{ ?}$$

(n goes to plus infinity.)

I tried putting in $n=1,2,3,4,\ldots$ to find some pattern but it's hard to see where it's going.

For example, $n=1$, limit is $\dfrac{1}{\sqrt{2}}$

For example, $n=2$, limit is $\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{6}}$

For example, $n=3$, limit is $\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{6}} + \dfrac{1}{\sqrt{12}}$

For example, $n=4$, limit is $\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{6}} + \dfrac{1}{\sqrt{12}} + \dfrac{1}{\sqrt{20}}$

I'm not exactly sure what this limit is converging to...

• Even though you've got a good answer, you might take another look at the question to make sure you really understand what is being asked. For example, when $n=2$ the correct expression is ${1\over\sqrt{5}}+{1\over \sqrt{6}}$, not ${1\over\sqrt{2}}+{1\over \sqrt{6}}$. – user940 Oct 5 '14 at 21:35
• @ByronSchmuland Wow... You're right. n is fixed for all terms, rather changing from 1 to n for every term... – bodacydo Oct 5 '14 at 21:51
• Duplicate of this – Yuriy S Aug 1 '16 at 17:03

Hint:

$$\frac{n}{\sqrt{n^2+n}}\leq(\frac{1}{\sqrt{n^2+1}} + ... + \frac{1}{\sqrt{n^2+n}}) \leq \frac{n}{\sqrt{n^2+1}}$$

• @bodacydo Essentially I used $\frac{1}{\sqrt{n^2 + n}} \leq \frac{1}{\sqrt{n^2 + i}} \leq \frac{1}{\sqrt{n^2 + 1}}$ n times – Petite Etincelle Oct 5 '14 at 21:23
• Ah thanks! I solved it. :) Answer is 1 :)) – bodacydo Oct 5 '14 at 21:30
• @bodacydo Yes indeed :) – Petite Etincelle Oct 5 '14 at 21:31

I want to propose a more powerful method which might be of your use in harder problems. That is use of the Riemann definition for the integral. For your question, we could write $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{1}{{\sqrt {{n^2} + i} }}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{1}{{\sqrt {1 + \frac{i}{n}\frac{1}{n}} }}\frac{1}{n}}$$now, lets assume that this is a discritized version of a continuous integral, for that assume that $x = \frac{i}{n}$, clearly in this case $dx$ which is the difference between two "consecutive" values of $x$ is ${\frac{1}{n}}$ so that we get $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{1}{{\sqrt {{n^2} + i} }}} = \int\limits_0^1 {\frac{{dx}}{{\sqrt {1 + xdx} }}} = \int\limits_0^1 {dx}$$Okay, I get that this is like using a cannon for a fly, but things would have got more interesting if the $i$ in the denominator of your question was also squared. Hope it helps ;)

• Thanks, but this is a bit way over my understanding :) For example, how did you go in the last equation to integral from 0 to 1, rather than from 1 to infinity? Also... how did you end up with $$sqrt(1 + x*dx)$$ ... Also how did nominator and denominator cancel to just have $dx$... – bodacydo Oct 5 '14 at 21:42
• @bodacydo For the first part of your question, place the limits of the summation $i=0$ and $i=n$ in the definition of $x = \frac{i}{n}$ and you would get the limits of the integral. But for the second part, I don't understand what you are asking. Could you clarify what is ambitious? – Seyed Mohsen Ayyoubzadeh Oct 5 '14 at 21:46
• Thanks, the first part is now clear. The second part ... it's not quite clear how you went from $$\sqrt{n^2+i}$$ in the denominator to $$\sqrt{1+xdx}$$ ... and it's also not clear how you went from $$\frac{dx}{\sqrt{1+xdx}}$$ to just $$dx$$. Did I make myself clearer? – bodacydo Oct 5 '14 at 21:48
• @bodacydo ok, factor ${n^2}$ from under the square root sign to get the R.H.S. of the first equation in my post. Now, you should be able to see both $x$ and $dx$ (don't you?). And then the factored part makes a $\frac{1}{n}$ for itself which is the $dx$ which has appeared. I hope it has helped. – Seyed Mohsen Ayyoubzadeh Oct 5 '14 at 21:53
• @bodacydo and for the second part, note that $$\frac{{dx}}{{\sqrt {1 + xdx} }} = \left( {\left( {1 - \frac{{xdx}}{2}} \right) + O(d{x^2})} \right)dx = dx$$the rule of the thumb is neglecting all $dx$s besides the one in the front of the integration sign (note that $\int {O(d{x^2})} = 0$) – Seyed Mohsen Ayyoubzadeh Oct 5 '14 at 21:57