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I'm getting crazy with the next limit: $$\lim_{n\rightarrow\infty}\frac{\sum \limits_{k=1}^n(k·a_k)}{n^2}$$ The exercise also says that is known that: $$\lim_{n\rightarrow\infty}a_n = L$$

I suppose that I have to answer in function of "$L$", but I still can't see anything. Any ideas?

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If $a_k=L$ for all $k$ then $s_k$ (the new sequence) converges to $L/2$. Now, use $\varepsilon$ to prove that it's indeed the result. –  Davide Giraudo Jan 15 '12 at 20:08
    
@Srivatsan Thanks for the hint, I could solve it and the result, as Davide Giraudo said, the result is L/2 –  Alejandro Jan 15 '12 at 20:40

2 Answers 2

up vote 4 down vote accepted

There are a few ways of doing this problem - ranging from first principles to using the Stolz-Cesàro theorem as a canned theorem. I will explain both the proofs here; however, note that if we unroll the proof of Stolz-Cesàro, the resulting proof is essentially the same as the second approach.

Method 1: Using the Stolz-Cesàro theorem: The Stolz-Cesàro theorem can be seen as a discrete version of l'Hôpital's rule. Verify that the denominator $n^2$ satisfies the hypotheses of the theorem; i.e., is strictly increasing and unbounded. Now $$ \frac{\sum \limits_{k=1}^n k a_k - \sum \limits_{k=1}^{n-1} k a_k}{n^2 - (n-1)^2} = \frac{n a_n}{2n-1} = \frac{a_n}{2 - \frac{1}{n}} \to \frac{L}{2} $$ as $n \to \infty$. Therefore the given sequence also converges to the same limit $\frac{L}{2}$. $\quad \diamond$


Method 2: First principles approach. This is a fleshed out version of Davide's hint. Fix $\varepsilon > 0$. Since $a_n \to L$ as $n \to \infty$, there exists $q$ such that $|a_n - L| \leqslant \varepsilon$ for all $n \geqslant q$. Therefore, $$ \begin{align*} \left| \frac{1}{n^2} \sum_{k=1}^n k a_k - \frac{L}{2} \right| &= \left| \frac{1}{n^2} \sum_{k=1}^n k (a_k - L) + \frac{1}{n^2} \sum_{k=1}^n k L - \frac{L}{2} \right| \\ &= \left| \frac{1}{n^2} \sum_{k=1}^n k \cdot (a_k - L) + \frac{L}{2n} \right| \\ &\leqslant \frac{1}{n^2} \sum_{k=1}^n k \cdot |a_k - L| + \frac{L}{2n} \\ &\leqslant \frac{1}{n^2} \sum_{k< q} k \cdot |a_k - L| + \frac{1}{n^2} \sum_{k=q}^n k \cdot \varepsilon + \frac{L}{2n} \\ &\leqslant \frac{1}{n^2} \sum_{k< q} k \cdot |a_k - L| + \varepsilon + \frac{L}{2n} . \end{align*} $$ Now for fixing $\varepsilon$ and $q$, for large enough $n$, we can bound the above by $2 \varepsilon$. Since this holds for all $\varepsilon > 0$, it follows that the sequence approaches $\frac{L}{2}$. $\quad \diamond$


Note: Connection to a generalised Cesàro theorem. The following theorem generalises the standard theorem on Cesàro means.

Suppose $\sum_{n} b_n$ is a positive series that diverges to infinity. Then if $(a_n) \to L$, then the generalised Cesàro mean $$ \frac{\sum_{k=1}^{n} b_k \cdot a_k}{\sum_{k=1}^{n} b_k} $$ also converges to $L$.

(The term “generalised Cesàro mean” is something I made up just now; I do not know if it is standard or not.) This theorem can be proved by mimicking the above two approaches, so I will skip the proof.

Applying the above theorem to $b_n = n$, we conclude that $$ \frac{\sum_{k=1}^{n} k \cdot a_k}{\sum_{k=1}^{n} k} = 2 \frac{\sum_{k=1}^{n} k \cdot a_k}{n(n+1)} $$ converges to $L$. From this, it follows that $\frac{\sum_{k=1}^{n} k \cdot a_k}{n^2}$ converges to $\frac{L}{2}$.

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Nice, just a little thing, shouldnt be?: $$\lim_{n\rightarrow\infty}\frac{\sum_{k=1}^{n+1}(k·a_k)-\sum_{k=1}^n(k·a_k)}{(n‌​+1)^2-n^2}$$ the Stolz formula is $\frac{a_{n+1}+a_n}{b_{n+1}+b_n}$ –  Alejandro Jan 16 '12 at 3:36
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@Alejandro Since we are considering asymptotics, $n$ vs. $n+1$ does not matter: we can shift the sequence by $1$ (to the left or to the right) and the limit does not change. In fact, the formula can equivalently be stated as $\frac{a_{n+10} - a_{n+9}}{b_{n+10} - b_{n+9}}$. [You got the negative sign flipped in the formula.] –  Srivatsan Jan 16 '12 at 3:40
    
Whops, you are right I flipped the sign. Very informative, thanks for the aclaration. –  Alejandro Jan 16 '12 at 3:46
    
@AlejandroGarcia I have expanded my answer adding a second first-principles proof (suggested by Davide). Hope you find this helpful. –  Srivatsan Jan 16 '12 at 4:21

Hint: It will be $$\lim_{n\rightarrow\infty}\frac{\sum_{k=1}^n(kL)}{n^2}$$ which is $$L \lim_{n\rightarrow\infty}\frac{\sum_{k=1}^n k}{n^2}$$ which you should be able to work out.

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But L is the limit of the "n" element, not for every term –  Alejandro Jan 15 '12 at 20:24
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I mean, L is the limit of a single element, not all the summation –  Alejandro Jan 15 '12 at 20:41
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You use an iterated application of the convergence of Cesaro means. –  ncmathsadist Jan 15 '12 at 20:55

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