# $\sum_{k=1}^{\infty}(a_k+\epsilon/2^k)$, where $a_k\ge 0$

I know that if $\sum a_k$ and $\sum b_k$ converge, then:

$$\sum_{k=1}^{\infty}(a_k+b_k)=\sum_{k=1}^{\infty}a_k+\sum_{k=1}^{\infty}b_k$$

However, as I read books on measure theory, I come across lines like this all the time:

$$\sum_{k=1}^{\infty}(m^*(E_k)+\epsilon/2^k)=\sum_{k=1}^{\infty}m^*(E_k)+\sum_{k=1}^{\infty}\epsilon/2^k$$

They are not the same concept, similar, but not the same as in the second case $\sum_{k=1}^{\infty}m^*(E_k)$ does not necessarily converge. At this stage of the proof, we do know that $0\le m^*(E_k)<+\infty$.

So, can I ask someone to provide a rigorous proof of the following statement:

If $a_k\ge 0$ for all $k$, then:

$$\sum_{k=1}^{\infty}(a_k+\epsilon/2^k)=\sum_{k=1}^{\infty}a_k+\epsilon$$

Note, I understand that $\sum_{k=1}^{\infty}\epsilon/2^k=\epsilon$, so you don't have to worry about showing that.

Thanks.

If $\sum{a_k}$ does not converge, it is $+\infty$ by your assumption that $a_k\ge 0$ for all $k$, and therefore the right hand side is $+\infty$. But then the left hand side is also $+\infty$, by term by term comparison (since $a_k+\epsilon/2^k\ge a_k$ for all $k$).

The other possibility is that $\sum a_k$ converges, and then we can apply the result you quote. In both cases, we have equality.

The condition that $\sum\limits_k a_k$ and $\sum\limits_k b_k$ both converge ensures that the identity $$\sum_{k=1}^{\infty}(a_k+b_k)=\sum_{k=1}^{\infty}a_k+\sum_{k=1}^{\infty}b_k$$ holds. But a different condition ensures that this identity holds, which is that $(a_k)$ and $(b_k)$ are nonnegative (then the identity holds in $[0,+\infty]$, naturally).