How come $\sum_{k=0}^\infty \frac{2}{k}$ and $\sum_{k=0}^\infty (k+1) \div \frac{k^2 + k}{2}$ equal the same thing? $$\sum_{k=0}^\infty \frac{2}{k} \tag1$$
and
$$\sum_{k=0}^\infty (k+1) \div \frac{k^2 + k}{2} \tag2$$
Are equal to the same thing:
$$\begin{align}
\text{UNDEF} &+ 2 + 1 + 0.\overline{66} + 0.5 + 0.4 + 0.\overline{33} + 0.\overline{285714} + 0.25 \\ &+ 0.\overline{22} + 0.2 + 0.\overline{18} +  0.1\overline{66} +0.\overline{153846}+ 0.\overline{142857}+ \ldots
\end{align} \tag3$$
This is quite interesting. The sigma expression at the bottom is basically taking a value, and seeing how big it is in relation to the "additive factorial sum" of it the number below it. Now, additive factorial sum is not a proper term I think, but I used it to describe the sum you get when you put a number into a binomial coefficient like $\frac{n^2 + 2}{2}$. So if n is 3, the answer is 6, because $1+2+3 =6$. Now, $3+1 =4$ and I wondered, how does 4 stack up against 6, or how does n+1 stack up against $\frac{n^2 + 2}{2}$?
The second sigma expression answers this, and it happens to be the same as the fist one, weirdly. How come?
 A: $$(k+1)\div\frac{k^2+k}{2}=(k+1)\cdot \frac{2}{k^2+k}=\frac{2(k+1)}{k(k+1)}=\frac 2k$$
In other words, the summands are both the same, although both sums are divergent, so you can't say that the sums are equal.
Also, it should be summed from $1$ to $\infty$, as $\frac 2k$ and $(k+1)\div\frac{k^2+k}{2}$ are undefined at $k=0$. When doing infinite sums, it is okay for the overall sum to be $\infty$, but the summand should always be defined on the range given.
A: First sum is divergent (it's harmonic series), so you can't say about eqaulity in this case, only for finite sums $\sum_{k=1}^n \frac{2}{k}$ and $\sum_{k=1}^n(k+1) \div \frac{k^2 + k}{2}$.
Also, I suppose there was a mistake and $k$ starts from 1.
A: In this case you can manipulate $ (k+1) \div \frac{k^2 + k}{2}$ and get $2/k$ as Rhys Hughes did in his answer, but note that, even if that is not possible, two sequences don't need to be equal in order to have the same limit!
For example. $1/n$ and $\sin(n)/e^n$ are diferent for all $n$ but note that:
$$\lim \frac{1}{n} = \lim\frac {\sin(n)}{e^n} = 0$$
