Verified answer

The infinite sum is the limit of a sequence of partial sums. So this is easy if you already have the property that sequence A is (a_1, a_2, ...), sequence B is (b_1, b_2, ...) and each sequence approaches a limit as n->infinity, then the sequence (a_1*b_1, a_2*b_2, ..., a_n*b_n, ...) also has a limit:

(lim a_n) (lim b_n) = lim (a_n*b_n)

(read "_n" as "sub n", and let the limits be from n=1 to infinity.)

Since an infinite sum is defined as the limit of a sequence of partial sums,

s_n = Σ_n a_i = a_1 + a_2 + ... + a_n .... (Σ_n is sum from i=1 to n, here)

s = Σ s_n = lim/n->oo s_n .... (Σ is sum from 1 to infinity (oo) and lim/n->oo is limit as n->oo)

Then:

cs = c * lim/n->00 s_n .... mulitply both sides by c

= lim/n->oo cs_n .... apply product of sequences formula from above

But cs_n is c times the finite sum Σ_n a_i, so the distributive law applies and that is equal to:

cs = lim/n->oo Σ_n ca_i

So that limit exists, whenever s=Σa_i converges.

That the "if part", The "only if" part is valid only if c is nonzero. You can't conclude that Σa_i converges just because Σ 0*a_i does. So, presume c is nonzero and let c' = 1/c, a'_i = ca_i. Then if Σ ca_i converges, that means Σa'_i converges, which (from above) means that Σc'a'_i converges, and that just the sum Σa_i, so if Σca_i converges, so does Σa_i.