Let $\sum a_n$ be a series of positive real numbers, n going from 0 to $\infty$, such that the series DOES NOT converge. Let $\left\{z_n\right\}$, $n\geq 0$ be a sequence of complex numbers. Form an associated sequence $\left\{w_n\right\}$ where $$ w_n = \frac{\sum_{l=0}^n a_l z_l}{\sum_{l=0}^n a_l} $$ Then prove that:

  1. if the sequence $\left\{z_n\right\}$ is convergent to some complex number $a$, then $w_n$ also converges to a.
  2. if $\left\{z_n\right\}$ is a real sequence and $lim_{n\to\infty} z_n = +\infty$, then $lim_{n\to\infty} w_n = +\infty$, as well.

Sorry for the awkward typing. Please ask for any clarifications. Thanks a lot.

  • $\begingroup$ The usual epsilon-delta approach works nicely here. What have you tried? $\endgroup$ – Did Feb 20 '14 at 8:52
  • $\begingroup$ I couldn't start the question. $\endgroup$ – user130113 Feb 20 '14 at 8:59
  • $\begingroup$ Have you solved any other exercise using the epsilon-delta approach? $\endgroup$ – Did Feb 20 '14 at 9:01
  • $\begingroup$ Nope. Could you please guide me through the proof? I am really new with this stuff and still learning. $\endgroup$ – user130113 Feb 20 '14 at 9:04
  • $\begingroup$ Try the theorem of Cesaro-Stolz, the discrete version of l'Hopitals rule. $\endgroup$ – LutzL Feb 20 '14 at 9:07

Without invoking Cesaro-Stolz, "epsilon-delta" argument:

Suppose $z_n \to a$. Let $S_n = \sum_{k=0}^n a_k$. We have $$ \left|w_n - a\right|=\left|\frac{\sum_{k=0}^n a_k \left|z_k - a\right|}{\sum_{k=0}^n a_k}\right|\leq \frac{1}{S_n}\sum_{k=0}^n a_k \left|z_k - a\right| $$ Fix $\varepsilon > 0$. Since $z_k\to a$, there is $N$ such that for $n>N$ we have $\left|z_n - a\right|<\varepsilon/2$. Now,

$$ \left|w_n-a\right|\leq \frac{1}{S_n}\sum_{k=0}^N a_k \left|z_k - a\right|+ \frac{1}{S_n}\sum_{k=N+1}^n a_k \left|z_k - a\right| $$

But $S_n\to\infty$ by assumption (positive terms, does not converge), so the first part tends to $0$ (sum remains fixed). And $$ \frac{1}{S_n}\sum_{k=N+1}^n a_k \left|z_k - a\right|< \frac{1}{S_n}\sum_{k=N+1}^n a_k \frac{\varepsilon}{2}= \frac{\varepsilon}{2S_n}\sum_{k=N+1}^n a_k $$

Now, $$ \sum_{k=N+1}^n a_k = \sum_{k=N+1}^n a_k + \sum_{k=0}^N a_k - \sum_{k=0}^N a_k = \sum_{k=0}^n a_k - \sum_{k=0}^N a_k = S_n - S_N $$ and so $$ \frac{1}{S_n}\sum_{k=N+1}^n a_k=\frac{S_n - S_N}{S_n}=1-\frac{S_N}{S_n}<1 $$ therefore $$ \frac{\varepsilon}{2S_n}\sum_{k=N+1}^n a_k<\frac{\varepsilon}{2} $$ Hence, for large enough $n$, $\left|w_n-a\right|<\varepsilon$.

Suppose now $z_n\to+\infty$. Fix $M > 0$. From some point, $z_n > M + 1$. Similar argument shows, that part of sum before that point can be neglected (bounded in absolute value by $\varepsilon$), and after that it's big enough to guarantee that limit is greater than $M$:

$$ \frac{1}{S_n}\left|\sum_{k=0}^N a_k z_k\right|\to 0 $$ as $n\to \infty$, since $N$ remains fixed, and $S_n\to\infty$. For large enough $n$ it is thus bounded by $\varepsilon$. $$ \frac{1}{S_n}\sum_{k=N+1}^n a_k z_k \geq \frac{1}{S_n}\sum_{k=N+1}^n a_k (M+1)=\frac{S_n - S_N}{S_n}(M+1)>M+\varepsilon $$

eventually, since $S_N/S_n\to 0$. Finally,

$$ w_n=\frac{1}{S_n}\sum_{k=0}^n a_k z_k\geq \frac{1}{S_n}\sum_{k=N+1}^n a_k z_k-\frac{1}{S_n}\left|\sum_{k=0}^N a_k z_k\right|>M+\varepsilon-\varepsilon=M $$

Therefore, $w_n > M$ for large enough $n$.

  • $\begingroup$ This looks great. Thanks a lot. $\endgroup$ – user130113 Feb 20 '14 at 17:24
  • $\begingroup$ Could you please explain why in the 3rd statement of equations, you said that ONLY the first part tends to zero and the second part doesn't? $\endgroup$ – user130113 Feb 21 '14 at 1:54
  • $\begingroup$ The first part is something constant divided by $S_n$, so it obviously tends to 0. The second part does as well, it's just less obvious. $\endgroup$ – Marcin Łoś Feb 21 '14 at 1:57
  • $\begingroup$ ok, so ∑ a_k |z_k - a| = ∑ a_k (e/2) = constant? How, can you please elaborate? Also, how did you get (S_n - S_N)/S_n? Sorry for so many questions. It just doesn't seem very obvious to me. $\endgroup$ – user130113 Feb 21 '14 at 2:08
  • $\begingroup$ I've edited the response to include more details. And I don't mind the questions - they help to improve the answer, after all. Some steps were/are sketchy, so it's no wonder they might require clarification. $\endgroup$ – Marcin Łoś Feb 21 '14 at 2:23

By the theorem of Cesaro-Stolz, under the given assumptions and with any second series $\sum b_k$,

$\frac{\sum_{k=1}^nb_k}{\sum_{k=1}^na_k}$ converges to the limit of $\frac{b_n}{a_n}$,

if this second limit exists. In this case, with $b_n=a_nz_n$, the fraction reproduces the sequence $z_n=\frac{b_n}{a_n}$.


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