I'm getting crazy with the next limit:
limn→∞n∑k=1(k·ak)n2
The exercise also says that is known that: limn→∞an=L
I suppose that I have to answer in function of "L", but I still can't see anything. Any ideas?
Answer
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 n2 satisfies the hypotheses of the theorem; i.e., is strictly increasing and unbounded. Now
n∑k=1kak−n−1∑k=1kakn2−(n−1)2=nan2n−1=an2−1n→L2
as n→∞. Therefore the given sequence also converges to the same limit L2. ⋄
Method 2: First principles approach. This is a fleshed out version of Davide's hint. Fix ε>0. Since an→L as n→∞, there exists q such that |an−L|⩽ 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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