I have two questions about limits (these questions arise from a proof that can be found in Page 43 and 44 respectively).
I will try to make clear the assumptions of each question, since they are located in a quite complicated proof. That is, I will write all hypothesis that are in the proof.
First question:
Suppose that $v_m \rightarrow 0$.
How to prove that $$\lim_{M \rightarrow +\infty} \frac{\sum_{m=2}^M mv_m}{\sum_{m=1}^M m} = 0 \quad ?$$
UPDATE: Solved! (Using Stolz's Theorem) Please see next question below.
Second question:
How to prove that
$$\lim_{n \rightarrow +\infty} \frac{\sum_{i=1}^{b_n}\alpha_i}{\log_2(n)}=0$$
given that $$\lim_{m \rightarrow +\infty}\alpha_m = 0$$ and $$b_n \leq p\left( \frac{\log_2(n)}{\log_2(p)}+1 \right),$$ where $p$ is a prime number (constant).
Suggestion given in the book: the numerator has at most $o(\log_2(n))$ terms $\alpha_i$. (I think that this is false, since the quotient between the numerator and $\log_2(n)$ is not zero...)
Thanks in advance
Clearly $\frac {b_n} {log_2 (n)}$ is bounded. Also $\lim \alpha_m =0$ implies that $\frac {\sum_1 ^{N} \alpha_i} {N} \to 0$ as $N \to \infty$. Let $|\frac {\sum_1 ^{N} \alpha_i} {N}| < \epsilon$ for $N \geq m$. If $b_n >m$ then $\frac {\sum_1 ^{b_n} \alpha_i} {b_n} < \epsilon$. Now multiply and divide by $\log_2 (n)$ and use the boundedness of $\frac {b_n} {log_2 (n)}$. Finally, to handle the case $b_n \leq m$ simply note that the numerator remains bounded and the denominator $\log_2 (n) \to \infty$. [ The numerator can take only the values $\alpha_1$, $\alpha_1+\alpha_2$,...,$\alpha_1+\alpha_2+\cdots+\alpha_m$ so it has a bound independent of $n$].