If $a_k>0\ \forall\ k\in \mathbb{N}\ $ and $r>0,\ $ prove that $\sum\frac{1}{a_k}$ converges $\iff \sum\frac{1}{a_k+r}$ converges.

Question

If $a_k > 0\ \forall\ k \in \mathbb{N}\ $ and $r>0,\ $ prove that $\displaystyle \sum_{k=1}^\infty\frac{1}{a_k}$ converges $\iff \displaystyle \sum_{k=1}^\infty\frac{1}{a_k+r}$ converges.

$\implies\ $ is easy: $\frac{1}{a_k + r}<\frac{1}{a_k}\ \forall\ k \in \mathbb{N}.$

$\impliedby\ $ is more difficult. I provide my answer below, but certainly welcome other answers: there are often simpler solutions as I tend to over-complicate things.

Answer 1

We may use the limit comparison test. Note that, for either series to converge, we must have $$\frac{1}{a_k}\to 0\Leftrightarrow a_k\to\infty.$$ In particular, for all $N$, there exists some $k_0$ for which $a_k>N$ for all $k>k_0$. We claim that $$\lim_{k\to\infty}\frac{a_k}{a_k+r}=1.$$ Indeed, for any $\epsilon$, $$\left|1-\frac{a_k}{a_k+r}\right|<\epsilon$$ is equivalent to $a_k>N$ for some $N$, so there exists a $k_0$ for which this holds for all $k>k_0$. By the limit comparison test, the convergences of the two series are equivalent.

Answer 2

We are trying to prove that, if $r>0,\ $ $\large{\frac{1}{a_k-r}}$ $\ > 0\ \forall\ k \in \mathbb{N}\ $ and $\ \displaystyle \sum_{k=1}^\infty\frac{1}{a_k}\ $ converges then $\displaystyle \sum_{k=1}^\infty\frac{1}{a_k-r}\ $ also converges.

I will prove the contrapositive of $\impliedby,\ $ that is, $\displaystyle \sum_{k=1}^\infty\frac{1}{a_k}\ $ diverges $\implies\ \displaystyle \sum_{k=1}^\infty\frac{1}{a_k+r}\ $ diverges.

If $\displaystyle\lim_{k\to\infty}\left(\frac{1}{a_k}\right) \neq 0,\ $ then $\exists\ c>0\ $ such that $\large{\frac{1}{a_k}}$ $ \geq c\ $ for infinitely many $k,\ $ which implies $\large{\frac{1}{a_k+r} \geq \frac{c}{1+rc}}$ $>0\ \forall\ k \in \mathbb{N},\ $ which implies that $\displaystyle \sum_{k=1}^\infty\frac{1}{a_k+r}\ $ diverges, and we are done. So let's assume for the rest of the proof that $\displaystyle\lim_{k\to\infty}\left(\frac{1}{a_k}\right) = 0.$

Then $\exists\ j \in \mathbb{N}\ $ such that $a_k \geq r\ \forall\ k \geq j,\ $

\begin{align} \implies\ \frac{1}{2a_k} = \frac{1}{a_k + a_k} \leq \frac{1}{a_k + r}\ \forall k \geq j\\ \\ \implies \displaystyle \sum_{k=1}^\infty\frac{1}{a_k+r} = \displaystyle \sum_{k=1}^{j-1}\frac{1}{a_k+r}+\displaystyle \sum_{k=j}^\infty\frac{1}{a_k+r} > \displaystyle \sum_{k=j}^\infty\frac{1}{a_k+r} \geq \frac{1}{2} \displaystyle \sum_{k=j}^\infty\frac{1}{a_k} = \infty.\\ \end{align}

Answer 3

We should restrict to the case that $a_{k}>0$ for all $k$. Otherwise, it is possible that $\sum_{k=1}^{\infty}\frac{1}{a_{k}+r}$ converges but $\sum_{k=1}^{\infty}\frac{1}{a_{k}}$ is undefined (because $\frac{1}{0}$ is undefined).

Since $\frac{1}{a_{k}+r}\leq\frac{1}{a_{k}},$ we have $\sum_{k=1}^{\infty}\frac{1}{a_{k}+r}\leq\sum_{k=1}^{\infty}\frac{1}{a_{k}}$. It follows that $\sum_{k=1}^{\infty}\frac{1}{a_{k}}$ converges $\Rightarrow$ $\sum_{k=1}^{\infty}\frac{1}{a_{k}+r}$ converges.

Conversely, assume that $\sum_{k=1}^{\infty}\frac{1}{a_{k}+r}$ converges. In particular, $\frac{1}{a_{k}+r}\rightarrow0$ and hence $a_{k}\rightarrow\infty.$ Now, $\frac{a_{k}+r}{a_{k}}\rightarrow1$, so there exists $K$ such that $\frac{a_{k}+r}{a_{k}}\leq2$ whenever $k\geq K$. Observe that \begin{eqnarray*} \sum_{k=K}^{\infty}\frac{1}{a_{k}} & = & \sum_{k=K}^{\infty}\frac{1}{a_{k}+r}\cdot\frac{a_{k}+r}{a_{k}}\\ & \leq & \sum_{k=K}^{\infty}\frac{1}{a_{k}+r}\cdot2\\ & < & \infty. \end{eqnarray*} It follows that $\sum_{k=1}^{\infty}\frac{1}{a_k}$ converges.