6

Given

$$ a_n >0 \text{ and } \lim_{n \to + \infty} a_n \left( \sum_{k=1}^n a_k \right) =2$$

I need to show that

$$\lim_{n \to + \infty} \sqrt{n}\ a_n=1$$

I tried first to compute $$\lim_{n \to + \infty} a_n,$$

but I don't know how, or how to handle these kind of questions so I appreciate any help.

Silverfish
  • 1,620
  • 21
  • 33
Butterfly
  • 1,443
  • Why isn't the limit of $a_n$ zero? It is. – Clement C. Dec 06 '15 at 20:44
  • i thought that being strictly positive they want us to see that it's not zero , how can you tell that it is ? because if it's not i won't have a limit =2 ? maybe i am mixing everything here – Butterfly Dec 06 '15 at 20:46
  • $(x_n)_n$ defined by $x_n=1/n$ is a strictly positive sequence, yet it converges to zero. Here, you are asked to show that $\sqrt{n}a_n \to 1$, which implies that $a_n \to 0$. – Clement C. Dec 06 '15 at 20:47
  • so if i wanted to prove that $a_n$ tends to zero before using that $\sqrt{n}a_n \to 1$ how can i do it ? and knowing that , how can i use it in order to prove what i am asked for ? – Butterfly Dec 06 '15 at 20:53
  • By contradiction, suppose not. Then the series $\sum_n a_n$ is not convergent (why?). But as this is a series of non-negative terms, this means $A_N=\sum_{n=1}^N a_n \xrightarrow[N\to\infty]{} \infty$. But since $a_n A_n \xrightarrow[n\to\infty]{} 2$, we then get that $a_n\sim_{n\to\infty} \frac{2}{A_n} \xrightarrow[n\to\infty]{} 0$, contradicting the assumption. So $a_n\xrightarrow[n\to\infty]{} 0$. – Clement C. Dec 06 '15 at 20:54
  • i see, i was mixing everything up , sry for that ! i appreciate your help, and i will try to finish it :) – Butterfly Dec 06 '15 at 20:59
  • @fleablood The sum is diverging. Since we want to show that $a_n\sim \frac{1}{\sqrt{n}}$, the sum $\sum_{k=1}^n a_n$ has to be divergent -- and equivalent to $\sum_{k=1}^n \frac{1}{\sqrt{n}}$. – Clement C. Dec 06 '15 at 21:37
  • oops. Didn't see that an in the summand. Changes things. Still. not specifying that a_n converges is not a valid reason to assume a_n diverges. In fact, specifying a_n is positive makes it less likely to diverge if a variety of sum converges (as it can't have converging differences as the values themselves diverge). – fleablood Dec 06 '15 at 21:44
  • I don't understand the point you are making, to be honest. What are you discussing/commenting on? – Clement C. Dec 06 '15 at 21:57

2 Answers2

5

Let $S_n=a_1+\cdots+a_n$. Then $T_n=S_n^2-S_{n-1}^2=S_n^2-(S_n-a_n)^2=2a_n S_n-a_n^2$. Hence as $a_n\to 0$ by @Clement C comment, $T_n\to 4$. By Cesaro's theorem, we get that $\displaystyle \frac{T_1+\cdots+T_n}{n} \to 4$, hence $\frac{S_n^2}{n}\to 4$, and it is easy to conclude.

Kelenner
  • 18,734
  • 26
  • 36
3

  • First, we do have $a_n\xrightarrow[n\to\infty]{} 0$: indeed, by contradiction, suppose not. Then the series $\sum_n a_n$ is not convergent (why?). But as this is a series of non-negative terms, this means $A_N=\sum_{n=1}^N a_n \xrightarrow[N\to\infty]{} \infty$. But since $a_n A_n \xrightarrow[n\to\infty]{} 2$, we then get that $a_n\sim_{n\to\infty} \frac{2}{A_n} \xrightarrow[n\to\infty]{} 0$, contradicting the assumption.

  • Now, write $A_N = \sum_{n=1}^N a_n$ with the convention $A_0=0$. By summing the above, the fact that $a_n A_n \xrightarrow[n\to\infty]{} 2$, and properties of divergent series, we have that $$ \sum_{n=1}^N a_n A_n \operatorname*{\sim}_{n\to\infty} 2n. $$ and similarly $$ \sum_{n=1}^N a_{n+1} A_n \operatorname*{\sim}_{n\to\infty} 2n. $$ (the latter as $a_{n+1} A_n = a_{n+1} A_{n+1} - a^2_{n+1} \xrightarrow[n\to\infty]{} 2$ as well.) We can now rewrite this as $$\begin{align*} \sum_{n=1}^N a_n A_n &= \sum_{n=1}^N (A_n-A_{n-1}) A_n = \sum_{n=1}^N A_n A_n - \sum_{n=1}^N A_{n-1} A_n = \sum_{n=1}^N A_n A_n - \sum_{n=0}^{N-1} A_{n} A_{n+1} \\ &= A_N^2 + \sum_{n=1}^{N-1} A_n A_n - \sum_{n=1}^{N-1} A_{n} A_{n+1} = A_N^2 + \sum_{n=1}^{N-1} A_n(A_n-A_{n+1}) = A_N^2 - \sum_{n=1}^{N-1} A_n a_{n+1} \end{align*}$$ and rearranging, $$ A_N^2 = \sum_{n=1}^{N-1} A_n a_{n+1} + \sum_{n=1}^N a_n A_n \operatorname*{\sim}_{n\to\infty} 4n. $$ This implies $A_N\operatorname*{\sim}_{n\to\infty} 2\sqrt{n}$, and (finally!) $$ a_n\operatorname*{\sim}_{n\to\infty} \frac{2}{2\sqrt{n}} = \frac{1}{\sqrt{n}}$$ i.e. $$ \sqrt{n} a_n \xrightarrow[n\to\infty]{}1. $$

Clement C.
  • 67,323