Proving whether the difference of two series is convergent of divergent Determine whether the difference of the following two series is convergent or not and Prove your answer$$
\sum_{n=1}^\infty \frac{1}{n}  $$  and $$\sum_{n=1}^\infty \frac{1}{2n-1} $$
What i tried. I said that the difference of the two series is divergent. My proof is as follows. Find the difference of the two series to get $$\sum_{n=1}^\infty \frac{1}{n} -\sum_{n=1}^\infty \frac{1}{2n-1} = \sum_{n=1}^\infty \frac{n-1}{n(2n-1)}$$
But this diffcult to prove directly that $\sum_{n=1}^\infty \frac{n-1}{n(2n-1)}$ is divergent. So i tired proving by contradiction by assuming that it is convergent and by rearraging the above equation we have,$\sum_{n=1}^\infty \frac{1}{n} =\sum_{n=1}^\infty \frac{1}{2n-1} + \sum_{n=1}^\infty \frac{n-1}{n(2n-1)}$ and since $\sum_{n-1}^\infty \frac{n-1}{n(2n-1)}$ is convergent by our assumption and $\sum_{n=1}^\infty \frac{1}{2n-1}$ is also convergent (need to be proven) then the sum of both series also have to be convergent and thus contradicting the fact that $\sum_{n=1}^\infty \frac{1}{n}$ is divergent and thus proving the statement. Is my proof correct and is there a better prove. Could anyone explain the Prove to me. Thanks
 A: A few things need correction in your proof. First, let's compliment you for the idea of trying the contrapositive. Unfortunately, the idea falls flat due to a mistake:

$\sum_{n=1}^\infty \frac{1}{2n-1}$ is not convergent!

To do the question, we need an observation about the difference. First, write down the first few terms of the sum:
$$
\sum_{n=1}^\infty \left(\frac 1{2n-1}\right) = 1 + \frac 13 + \frac 15 + \frac 17 + ...
$$
you can see that we are precisely summing over the reciprocals of odd numbers here. 
In the series $\sum \frac 1n$, we are summing over reciprocals of all numbers. Hence, the difference of these two series, is the series whose terms are the reciprocals of all even numbers. This should be clear.
Once this is clear, we see that every even number is of the form $2k$, where $k$ is a natural number, and vice versa, so that:
$$
\sum \frac 1n - \sum \frac 1{2n- 1} = \sum \frac{1}{2n}
$$ 
(Note that the above has nothing to do with convergence of any series, since we are adding series it is mathematically sound).
Now, note that $\sum \frac{1}{2n}$ cannot converge, for if it did, say the sum is $M$, then $\sum \frac 1n = 2M$,  which is a contradiction as this series doesn't converge.
Hence, the difference does not converge.
To see that the reciprocals of odd numbers does not converge, just do:
$$
1 + \frac 13 + \frac 15 + \frac 17...  > \frac 12 + \frac 14 + \frac 16 + \frac 18...
$$
(Compare term by term)
Hence, we  have that by the comparison test this is also not convergent.

EDIT : As  the below comments point out, a rearrangement of the series $\sum \frac 1n - \sum \frac1{2n-1}$ may converge. However, the original problem also does not point this out explicitly, and in that case it is with me to decide which rearrangement is being referred to : and from the computations made above (where the difference was taken explicitly), I concluded that there was no rearrangement of the terms done while taking the difference. Nevertheless, this is a non-trivial thing that was not pointed out in the question.
A: For a direct comparison test, note that
$$
\frac{(n-1)}{n(2n - 1)} \geq \frac{n-1}{n(2n)} = \frac{n-1}{2n^2}
$$
If the sum converged, we would have
$$
\sum_{n=1}^\infty \frac{n-1}{2n^2} = 
\sum_{n=1}^\infty \frac{n}{2n^2} - 
\sum_{n=1}^\infty \frac{1}{2n^2}
$$
A: HINT: Observe that $$\sum_{n=1}^\infty\frac{n-1}{n(2n-1)}\ge\sum_{n=1}^\infty\frac{n-1}{n(2n)}=\sum_{n=1}^\infty\frac{1}{2n}-\sum_{n=1}^\infty\frac{1}{2n^2}.$$
A: The question is not meaningful: The difference of the series is not defined since one (actually, each one) of them is not defined.
If, nevertheless, you insist on going ahead and "subtracting" them, then why not try it in the following artistic way? $$\sum_{n=1}^\infty \frac{1}{n}-\sum_{n=1}^\infty \frac{1}{2n-1}=$$$$\frac11-\left(\frac11\right)+\frac12-\left(\frac13+\frac15\right)+\frac13-\left(\frac17+\frac19+\frac1{11}\right)+
\frac14-\left(\frac1{13}+\frac1{15}+\frac1{17}+\frac1{19}\right)+\cdots,$$or, more formally,$$\sum_{n=1}^\infty \frac{1}{n}-\sum_{n=1}^\infty \frac{1}{2n-1}=\sum_{n=1}^\infty\left(\frac{1}{n}-\sum_{k=(n^2-n)/2+1}^{(n^2+n)/2} \frac{1}{2k-1}\right).$$The series on the right-hand side is well defined. It converges quite nicely to a smallish negative quantity.
