# Series divergence - how precise should the answer be

Morning.

I've written down some of my reasoning and arguments as to why the series diverges, however I am not certain I can safely conclude it diverges to $\infty$. Would you give it a look, please?

$$\sum_{n=1}^{\infty} \frac{1+\dfrac{1}{n}}{n} = \sum_{n=1}^{\infty} \frac{1}{n} + \sum_{n=1}^{\infty} \frac{1}{n^2}$$

Thus by recognising the $\zeta(n)$ when $n=1$ we know $\dfrac{1}{n}$ diverges,
as a sum of an infinite amount of $\dfrac{1}{2}$ results in $\infty$.

Hence, anything added to $\infty$ must also be $\infty$, assuming it is defined. So lets check if the last sum is defined.

By the p-test it is, and hence converges ($p>1$). So, divergence + convergence = divergence.

Could I however say $\infty + a|a\in\mathbb Q = \infty$, that is a given that it is any defined number.

And hence say that this sum diverges TO $\infty$?

• Yes, you can say the sum diverges to infinity. People also may say it converges to infinity, but that is more confusing for beginners. Dec 15, 2013 at 13:27

There's no need to argue that $\infty + a = \infty$, since we can't literally perform "normal" arithmetic with "infinity", per se, though the idea behind it is sound. A divergent series grows without bound, so the sum of a divergent series and a convergent series is necessarily divergent, since adding a finite sum to an unbounded (infinite) sum is necessarily unbounded: it diverges to infinity.
So in the end, yes, you can certainly argue that $$\sum_{n=1}^{\infty} \frac{1+\dfrac{1}{n}}{n} = \left(\sum_{n=1}^{\infty} \frac{1}{n} + \sum_{n=1}^{\infty} \frac{1}{n^2}\right) = \infty$$
• I would end with "${}= \infty$", not "${}\to \infty$" Dec 15, 2013 at 13:22
• Yes, @GEdgar (I get so in the habit, in terms of computing limits, to writing $\to \infty$ Dec 15, 2013 at 13:24