# 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

## 1 Answer

After the discussion above the answer to my question is:

We are allowed to interchange summation and integration by the use of Tonelli's Theorem because that thing that lies in the series is positive.

And everything is going its way.

Thanks for the help!!