# Why is the interchange of integration and summation allowed in this case?

In a solution of a book of the integral:

$$\int_a^{\infty} \sum_{n=1}^{\infty} \frac{1}{(z+n)^{k+1}}\,dz, \;\; a\geq 1$$

I see the following:

\begin{align*} \int_{a}^{\infty}\sum_{n=1}^{\infty}\frac{1}{\left ( n+z \right )^{k+1}}\,dz &= \sum_{n=1}^{\infty}\int_{a}^{\infty}\frac{dz}{(n+z)^{k+1}}\\ &= \cdots\\ \end{align*}

The rest of the solution is understable to me but not the interchange. I was unable to prove that the fuction within the series converges uniformly... and I cannot think of something else that works here e.g monotone convergance thoerem or Tonelli Theorem.

• I think you have a typo and intended $dz$ instead of $dx$. I'm also confused about the roles of $n$ and $k$ between your two lines. Anyway, Fubini-Tonelli applied to the product of the Lebesgue and counting measures implies that you only need to prove that when you put absolute values inside the sum, you get a finite solution. – Ian Feb 18 '15 at 22:42
• Fixed for $dz$. Well $n$ is the index of the series and $k$ is just any fixed natural number. Suppose that $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(z+k)^{4}}$ meaning that $k=3$. – Tolaso Feb 18 '15 at 22:44
• My confusion is that in the first line you have $(z+k)^{n+1}$ while in the second line you have $(n+z)^{k+1}$. It is not clear which one you mean. – Ian Feb 18 '15 at 22:45
• I think it's ok.. got your point... I have just written it so quickly.. i did not check later for typos... Now, what do I have to prove? OK, I take absolute values and then what? – Tolaso Feb 18 '15 at 22:50
• Analysis is far from my strong suit, but why can't you use Tonelli's Theorem? – Callus - Reinstate Monica Feb 18 '15 at 22:56