1. p-series

p-series
The p-series is of the form
?_(k=1)^??1/k^p
converges if p>1, and
diverges if p?1.
Examples:
?_(k=1)^??1/k^2 converges since it is p-series with p=2.
?_(k=1)^??1/k diverges since it is a p-series with p=1.

Telescoping Series:
The series of the form
?_(k=1)^??1/(k(k+p))
where p is a natural number, is called telescoping series.
It involves two steps to study telescoping series:
First, we write the general term of the series as a difference of two fractions, using partial fraction decomposition.

Partial Fraction Decomposition:

1/(k(k+p))=1/kp-1/p(k+p) =1/p (1/k-1/((k+p) ))

Second, we find and simplify the sequence of partial sums, as most of its terms will cancel.


Convergence of Telescoping Series:
Instead of studying ?_(k=1)^??1/(k(k+p)), we study 1/p ?_(k=1)^??(1/k-1/((k+p) )) ,
since

1/p ?_(k=1)^??(1/k-1/((k+p) )) =?_(k=1)^???1/p (1/k-1/((k+p) )) ?


To study it, we use its sequence of partial sums, S_n. By definition,

S_n=?_(k=1)^n?(1/k-1/((k+p) ))

We first look at the special cases p = 1 and p = 2. We then generalize our result for any positive integer p.

Case1. p = 1
S_n=?_(k=1)^n?(1/k-1/((k+1) ))

S_n=(1-1/2)+(1/2-1/3)+(1/3-1/4)+?+(1/n-1/(n+1))=1-1/(n+1)=n/(n+1)

?lim??(n??)??S_n ?=1

Thus, the series ?_(k=1)^??1/k(k+1) converges and has the sum 1.

Case 2. p=2

The series ?_(k=1)^??1/k(k+2) =1/2 ?_(k=1)^??(1/k-1/((k+2) ))

S_n=1/2 ?_(k=1)^n?(1/k-1/((k+2) ))

S_n=1/2 [(1-1/3)+(1/2-1/4)+(1/3-1/5)+?(1/n-1/(n+1))+(1/n-1/(n+2))]
=1/2 [1+1/2-1/(n+1)-1/(n+2)]

?lim??(n??)??S_n ?=1/2 [1+1/2]

General Case

This is similar to above,

S_n=1/p ?_(k=1)^n?(1/k-1/((k+p) ))

S_n=1/p [(1-1/(1+p))+(1/2-1/(2+p))+(1/3-1/(3+p))+?+(1/n-1/(n+p))]
=1/p [1+1/2+1/3+?+1/p-1/(n+1)-1/(n+2)-…-1/(n+p)]

?lim??(n??)??S_n ?=[1+1/2+1/3+?+1/p]=1/p ?_(k=1)^p?1/k



Summary:
If p is any positive integer then
?_(k=1)^??1/k(k+p) =1/p ?_(k=1)^p?1/k

Examples:
The series
?_(k=1)^??5/k(k+3) =5/3 ?_(k=1)^3?1/k=5/3 (1+1/2+1/3)=55/18

The series
?_(k=1)^??1/k(k+5) =1/5 ?_(k=1)^5?1/k=1/5 (1+1/2+1/3+1/4+1/5)