Answer to Question #161759 in Calculus for Vishal

Question

Let ∞n=1an be a series of real numbers and N0 ∈ N. Show that ∞n=1an converges if and only if ∞n=N0 an converges

Answer 1

Let:

$\displaystyle\sum_{n=1}^\infin a_n=S_n$

then:

$\displaystyle\sum_{n=1}^\infin (N_0a_n)=N_0\cdot S_n$

If $\displaystyle\sum_{n=1}^\infin (N_0a_n)$ converges, then $(N_0\cdot S_n)\to S$ for some $S$, and $S_n\to S/N_0$ ; that is,

$\displaystyle\sum_{n=1}^\infin a_n$ converges.

If $\displaystyle\sum_{n=1}^\infin (N_0a_n)$ diverges, then $(N_0\cdot S_n)$ does not tend to some $S$ . So, $S_n$ also does not tend to some $(S/N_0)$ , that is, $\displaystyle\sum_{n=1}^\infin a_n$ diverges.