# Math Help - Proving the convergence of a series?

1. ## Proving the convergence of a series?

(a) Explain why, when proving the convergence of a series
∑ of a(n) between n=0 and ∞ it is permissible to ignore
finitely many terms of the series.

Let s(n)=∑ of a(n) between k=0 and n be the nth partial
sum. The series ∑ of a(n) between k=0 and ∞
converges iff the sequence (s(n)) of partial sums converges.

Now suppose that t(n)=∑ of a(n) between k=m and n
is the partial sum formed by ignoring the first m entries
and t(n)→l∈ℝ as n→∞.
Then s(n)=a(0)+a(1)+…+a(m-1)+t(n)
and s(n)→a(0)+a(1)+…+a(m-1)+l
by continuity of the operation of adding the constant value
a(0)+a(1)+…+a(m-1). So the
original series also converges.

(b) Use (a) together with a form of the comparison test to show
that the series ∑ of 1/(2n-5)(2n-3)(2n-3) between n=0 and ∞
converges.

Note: I have been told my first answer is correct, and could some also plz tell me who to post using the normal mathematical symbols, such as the sum of and a small n. Thankyou

2. Originally Posted by kbartlett
(a) Explain why, when proving the convergence of a series
∑ of a(n) between n=0 and ∞ it is permissible to ignore
finitely many terms of the series.

Let s(n)=∑ of a(n) between k=0 and n be the nth partial
sum. The series ∑ of a(n) between k=0 and ∞
converges iff the sequence (s(n)) of partial sums converges.

Now suppose that t(n)=∑ of a(n) between k=m and n
is the partial sum formed by ignoring the first m entries
and t(n)→l∈ℝ as n→∞.
Then s(n)=a(0)+a(1)+…+a(m-1)+t(n)
and s(n)→a(0)+a(1)+…+a(m-1)+l
by continuity of the operation of adding the constant value
a(0)+a(1)+…+a(m-1). So the
original series also converges.

(b) Use (a) together with a form of the comparison test to show
that the series ∑ of 1/ between n=0 and ∞
converges.

Note: I have been told my first answer is correct, and could some also plz tell me who to post using the normal mathematical symbols, such as the sum of and a small n. Thankyou

$\sum_{n=3}^{\infty}\frac{1}{(2n-5)(2n-3)(2n-3)} \le \sum_{n=3}^{\infty}\frac{1}{(2n)^3}$
this is a p-series with p=3. so it converges.