Given a diverging sequence, does another converge?

Question

Let $\{ a_n \} _{n=0}^{\infty}$ be a sequence of positive reals such that $\sum_{n=0}^{\infty} a_n$ diverges. Is it necessarily the case that $\sum_{n=0}^{\infty} \frac{a_n}{2016 + a_n}$ also diverges?

I split this into two cases based on whether $\{a_n\}$ has a subsequence that diverges to infinity. If a part of it diverges to infinity, say $\{b_n\}$, then $\lim_{n \to \infty} \frac{b_n}{2016 + b_n} = 1$ for infinitely many terms, and so it diverges.

I'm not sure about the other case, where $\{a_n\}$ does not have a subsequence diverging to infinity. I tried setting an upper bound, $k$, so that $a_i < k$ for all $i$, but I'm not sure how to continue.

Answer 1

$\{a_n\}$ does not have a subsequence diverging to infinity if and only if $(a_n)$ is bounded. If $a_n \leq M$ then $\frac {a_n}{2016+a_n} \geq \frac 1 N a_n$where $N=2106+M$.

Answer 2

Let $b_n = \frac{a_n}{2016 + a_n}$, note that $a_n = \frac{2016b_n}{1-b_n}$ (as $a_n>0$).

Then, it is clear that $a_n \underset{\infty}{\to}0$ iff. $b_n \underset{\infty}{\to}0$.

Then, your idea comes into play:

• either $a_n \underset{\infty}{\to}0$, then as $b_n \sim 2016a_n$, $\sum a_n$ and $\sum b_n$ both diverge (as $\sum a_n$ diverges)
• either $a_n \underset{\infty}{\not\to}0$, then $b_n \underset{\infty}{\not\to}0$ and $\sum b_n$ diverges.

Answer 3

For $a,b,c>0$ observe that $$ \frac{a}{a+c}+\frac{b}{b+c} > \frac{a+b}{a+b+c},\qquad (\star) $$ since $$ \frac{a}{a+c}+\frac{b}{b+c} - \frac{a+b}{a+b+c}=\bigl((a+c)+(b+c)\bigr)\left(\frac{1}{a+b+c}-\frac{1}{a+b+c+\frac{ab}{c}}\right). $$ By iteratively applying $(\star)$ to we find that the inequality extends to sequences as well, yielding $$ \frac{a_1}{a_1+c}+\frac{a_2}{a_2+c}+\cdots+\frac{a_n}{a_n+c}>\frac{a_1+\cdots+a_n}{c+a_1+\cdots+a_n}\qquad (\star\star). $$ Now consider a divergent series $\{a_n\}$. We group together subsequences each of which sums to $c$ or higher (which must exist by the divergence), which we notate as $$ a_{i_k}+\cdots+a_{j_k}\geq c,\qquad i_1<j_1<i_2<j_2<\cdots. $$ Applying $(\star\star)$ to each subsequence in turn yields $$ \sum_{n=i_k}^{j_k}\frac{a_n}{c+a_n}>\frac{c}{c+c}=\frac{1}{2}. $$ Thus the sum $$ \sum_{n=0}^{\infty}\frac{a_n}{c+a_n} $$ diverges, because it contains the sum over each of the subsequences, each contributing a constant $\tfrac12$ (or greater) to the sum. In particular when $c=2016$ this answers the question.