Prove that $b_n = \frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} + \dotsb + \frac{1}{\sqrt{n^2 + n}} \to 1$ as $n \to \infty$

Question

I need to prove that $b_n = \frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} + \dotsb + \frac{1}{\sqrt{n^2 + n}} \to 1$ as $n \to \infty$

My initial thoughts are to sandwich this (seems sensible to me). Finding the above inequality $b_n < 1$ was easy, since clearly $\frac{1}{\sqrt{n^2 + k}} < \frac{1}{n}$ for integer k. But I'm stuck on how to compare it with something from below.

A hint I've been given is that if $a_n \to a$, then $\sqrt{a_n} \to \sqrt{a}$ as $n \to \infty$, though I'm struggling to see a way of applying this.

score 4 · Accepted Answer · answered Apr 08 '15 at 09:02

As you can see,

$\frac{1}{\sqrt{n^2+1}}$ is the biggest term and $\frac{1}{\sqrt{n^2+n}}$ is the smallest one. Doing this, we can say:

$\frac{n}{\sqrt{n^2+n}}<b_n<\frac{n}{\sqrt{n^2+1}}$

Taking the limits on both sides of the inequality, we find that

$\lim_{n\to\infty}b_n=1$

QED.

Arpan · Answer 2 · 2015-04-09T03:34:02.677

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You can use $$\frac{1}{\sqrt{n^2+n}}\leq\frac{1}{\sqrt{n^2+k}}\leq\frac{1}{n}$$

edited Apr 09 '15 at 03:34

answered Apr 08 '15 at 08:59

Arpan

2,633

I like your integral approach! (Pun intended) – Hasan Saad Apr 08 '15 at 09:09
@HasanSaad Haha! That's the method that came to mind before I saw the OP approached by Squeeze theorem. – Arpan Apr 08 '15 at 09:10
It sure did not come to my mind... (Squeeze theorem seemed more "natural" in this case...) – Hasan Saad Apr 08 '15 at 09:14
Everyone thinks differently. I was able to visualize the $\frac{i}{n}$ terms easily. – Arpan Apr 08 '15 at 09:15
Yeah, I guess everyone thinks differently. – Hasan Saad Apr 08 '15 at 09:16
@Arpan Please, I don't understand your answer, why do you write $\displaystyle \sum_{i=1}^n\frac{1}{\sqrt{1+\left(\frac{i}{n}\right)^2}}$ whereas $$ \sqrt{n^2+i}=n\sqrt{1+\frac{i}{n^2}}\neq\sqrt{1+\left(\frac{i}{n}\right)^2}$$ – Euler's student Apr 08 '15 at 09:37
The $n$ is still there. – Arpan Apr 08 '15 at 09:39
@Arpan Yes, but there is no square for $i$... – Euler's student Apr 08 '15 at 11:05
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@Arpan You can observe that: $$ \int_0^1\frac{dx}{\sqrt{1+x^2}}=\ln(1+\sqrt{2})\neq 1$$ – Euler's student Apr 08 '15 at 16:38
@Euler'sstudent Right. Thanks for pointing that out! – Arpan Apr 09 '15 at 03:33

score 1 · Answer 3 · answered Apr 08 '15 at 09:01

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Hint. You can check that $$\sum_{k=1}^n\frac{1}{\sqrt{n^2+n}}\leq\sum_{k=1}^n\frac{1}{\sqrt{n^2+k}}\leq\sum_{k=1}^n\frac{1}{n}$$

answered Apr 08 '15 at 09:01

Euler's student

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score 1 · Answer 4 · answered Apr 08 '15 at 09:01

1

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

answered Apr 08 '15 at 09:01

Marco Cantarini

33,062
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@HasanSaad LOL! – Marco Cantarini Apr 08 '15 at 09:05

score 0 · Answer 5 · answered Apr 08 '15 at 08:59

0

Hint: which one of the summands is the smallest?

answered Apr 08 '15 at 08:59

Adayah

10,468

Prove that $b_n = \frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} + \dotsb + \frac{1}{\sqrt{n^2 + n}} \to 1$ as $n \to \infty$

5 Answers5