Does $\sum^{\infty}_{1}\frac{1}{k+1} = \sum^{\infty}_{2}\frac{1}{k}$ diverge because $\sum^{\infty}_{1}\frac{1}{k}$ diverges? If you manipulate the index of a series does it still converge/diverge?
For example:
Does $\sum^{\infty}_{1}\frac{1}{k+1} = \sum^{\infty}_{2}\frac{1}{k}$ diverge because $\sum^{\infty}_{1}\frac{1}{k}$ diverges?
 A: In general, for any fixed $n$:
$$\sum_{k=1}^{\infty} a_k=\sum_{k=1}^n a_k+\sum_{k=n+1}^{\infty}a_k,$$
if the last sum converges (diverges), then the first sum also converges (diverges).
A: $$
\sum_{k=1}^\infty \frac 1 {k+1} = \sum_{k=2}^\infty \frac 1 k = \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \cdots
$$
$$
\sum_{k=1}^\infty \frac 1 k = 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \cdots
$$
Changing only one term of a series, or finitely many terms, will alter the sum but will not alter whether it converges or diverges.
A: $$\sum^{\infty}_{1}\frac{1}{k+1}=\frac 12 +\frac13 +\frac 14+\ldots$$
$$\sum^{\infty}_{1}\frac{1}{k}=1+\frac 12 +\frac13 +\frac 14+\ldots$$
Hence $$\sum^{\infty}_{1}\frac{1}{k+1}=\sum^{\infty}_{1}\frac{1}{k}-1$$
Since $$\sum^{\infty}_{1}\frac{1}{k}\to \infty \implies \sum^{\infty}_{1}\frac{1}{k}-\color{blue}1 \to \infty \implies \sum^{\infty}_{1}\frac{1}{k+1 }\to \infty$$
A: Changing a finite number of terms will not alter whether it converges or not. For instance, $$ \sum_{n=1}^\infty \frac{1}{n^2} \text{ and } \sum_{n=2}^\infty \frac{1}{n^2} $$
both converge, just to a different value.
However, be careful. Changing the index can create poles causing the sum to be undefined. Take for example,
$$ \sum_{n=0}^\infty \frac{1}{n^2} $$
The whole sum is undefined, because $\frac{1}{0}$ is a term.
