Limit question how do i solve it

Question

How to find the limit of this sequence:

$$\lim_{n\to \infty} (1^2+2^2+3^2+...+n^2)^\frac{1}{n}$$

What I tried: I tried diving the whole equation by $n^2$ and taking that out the bracket and so everything inside the bracket tends to 0 except the last term which tends to 1. This will mean that the sequence tends to 1 but if you input n values for n=1,2,3 etc you will see that the sequence is increasing so not sure

Have you tried simplifying the expression inside the parentheses? — Arthur, Nov 19 '17 at 00:00
Yes if i use the formula for sum of $r^2$ from 1 to n you would get ${\frac{n(n+1)(2n+1)}{6}}$ all to the power of $\frac{1}{n}$ which seems to tend to 1 — user504498, Nov 19 '17 at 00:05
@imranfat well i can show it tends to 1 but the problem is if i input n=1,2,3...etc the sequence gets bigger and bigger so how does it tend to 1 e.g. for n=1 its 1, n=2 its 2.236... n=3 its 2.5... — user504498, Nov 19 '17 at 00:10
$$n^2\leq 1^2+\ldots+n^2 \leq n^2+\ldots + n^2 = n^3 $$ and $$\lim_{n\to +\infty} n^{k/n} = 1$$ for any $k>0$, in particular for $k=2$ and $k=3$. — Jack D'Aurizio, Nov 19 '17 at 03:47

score 1 · Answer 1 · answered Nov 19 '17 at 00:23

Via this post, your limit is also equal to $$ \lim_{n\to\infty}\frac{1^2+2^2+\dotsb+(n+1)^2}{1^2+2^2+\dotsb+n^2} =\lim_{n\to\infty}1+\frac{(n+1)^2}{1^2+2^2+\dotsb+n^2} $$ But by the hint given by DonAntonio, $(1^2+2^2+\dotsb+n^2)/3^{-1}n^3\to1$. You should be able to take it from here.

Dylan · Answer 2 · 2017-11-19T00:32:17.593

Suppose the limit exists

$$ L = \lim_{n\to\infty}\left(\frac{n(1+n)(1+2n)}{6}\right)^{1/n} $$

We can take the natural log of both sides to get

$$ \ln(L) = \lim_{n\to\infty} \frac{1}{n}\left(\ln n + \ln(1+n) + \ln(1+2n) - \ln 6 \right) $$

It remains to prove that $$ \lim_{n\to\infty} \frac{\ln n}{n} = \lim_{n\to\infty} \frac{\ln(1+n)}{n} = \lim_{n\to\infty} \frac{\ln(1+2n)}{n} = \lim_{n\to\infty} \frac{\ln 6}{n} = 0 $$

Which gives $\ln (L) = 0$ or $L = 1$

To answer your other question as to why the sequence seems to increase. The plot of $a_n$ vs. $n$ looks like this

You can see that it initially increases but decreases past $n = 3$. In fact, it decreases very slowly to $1$:

$$ a_{10} = 1.813619 $$ $$ a_{100} = 1.135779 $$ $$ a_{1000} = 1.019820 $$ $$ a_{10000} = 1.002657 $$

score 0 · Answer 3 · answered Nov 19 '17 at 00:07

0

$$1^2+2^2+\ldots+n^2=\frac{n(n+1)(2n+1)}6$$

You might also find useful the fact that

$$\lim_{n\to\infty}\sqrt[n]n=\lim_{n\to\infty}\sqrt[n]a=1\;,\;\;\forall\,a> 0$$

answered Nov 19 '17 at 00:07

DonAntonio

211,718
17
136
287

i can show it tends to 1 but the problem is if i input n=1,2,3...etc the sequence gets bigger and bigger so how does it tend to 1 e.g. for n=1 its 1, n=2 its 2.236... n=3 its 2.5... – user504498 Nov 19 '17 at 00:15
1

@user504498: if you can show that a sequence tends to $1$ then it is obvious that as the value of $n$ increases the values of sequence will start to get close to $1$. I wonder why you are getting confused with few initial values of $n$. Did you check the value of the sequence for $n=10000$? Moreover checking different values of $n$ is not the key. The limiting behavior of a sequence may be observable empirically only after $n$ exceeds a certain value. This is formalized in the $\epsilon, N$ definition of limit and if $\epsilon$ is small you may need a very large $N$. – Paramanand Singh Nov 19 '17 at 13:41

Limit question how do i solve it

3 Answers3