Infinite Series
Key Questions

Answer:
See below
Explanation:
There are different types of series, to what use different methods of evaluating
For example a converging geometric series:
#a+ ar + ar^2 + ar^3 + ... + ar^k = sum_(n=1) ^(k) ar^(n1) # where
#sum_(n=1) ^(k) ar^(n1) = (a(1r^k)) / (1r) # Assuming
#r < 1 # we can let#k to oo # for infinite series to be evaluated ...#lim_(k to oo ) sum_(n=1) ^(k) ar^(n1) = lim_(k to oo ) ( a(1r^k) )/(1k) # as
#k to oo # ,#r^k to 0 # as#r< 1 # #=> sum_(n=1) ^(oo) ar^(n1) = a/(1r) # but there are other series what can be approached with tricks!
Take
# 1/6 + 1/12 + 1/20 + 1/30 + ... # After consideration we can recognise this is the same as...
# (1/2  1/3 ) + ( 1/3  1/4) + (1/4  1/5 ) + ... # # (1/2 cancel( 1/3) ) + (cancel( 1/3) cancel( 1/4)) + (cancel(1/4) cancel( 1/5) ) + ... # #=1/2 # There are also other infinite series that you can remember, and may be able to prove, a like:
#e^x = 1 + x + x^2 /(2!) + x^3 / (3!) + ... = sum_(n=0) ^oo x^n / (n!) # There are many others, where there insist one set way of computing infinite series, there are many!

I believe that it is the same as an alternating series. If that is the case, then an oscillating series is a series of the form:
#sum_{n=0}^infty (1)^n b_n# , where#b_n ge 0# .For example, the alternating harmonic series
#sum_{n=1}^infty{(1)^n}/n# is a convergent alternating series.

Here is an example of a telescoping series
#sum_{n=1}^infty(1/n1/{n+1})#
#=(1/11/2)+(1/21/3)+(1/31/4)+cdots#
As you can see above, terms are shifted with some overlapping terms, which reminds us of a telescope. In order to find the sum, we will its partial sum#S_n# first.
#S_n=(1/11/2)+(1/21/3)+cdots+(1/n1/{n+1})#
by cancelling the overlapping terms,
#=11/{n+1}#
Hence, the sume of the infinite series can be found by
#sum_{n=1}^infty(1/n1/{n+1})=lim_{n to infty}S_n=lim_{n to infty}(11/{n+1})=1# 
It is very tough to answer such a general question, but I will give it a shot. Infinite series allow us to add up infinitely many terms, so it is suitable for representing something that keeps on going forever; for example, a geometric series can be used to find a fraction equivalent to any given repeating decimal such as:
#3.333...# by splitting into individual decimals,
#=3+0.3+0.03+0.003+cdots# by rewriting into a form of geometric series,
#=3+3(1/10)+3(1/10)^2+3(1/10)^3+cdots# by using the formula for the sum of geometric series,
#=3/{11/10}=10/3# The knowledge of geometric series helped us find the fraction fairly easily. I hope that this was helpful.
Questions
Tests of Convergence / Divergence

Geometric Series

Nth Term Test for Divergence of an Infinite Series

Direct Comparison Test for Convergence of an Infinite Series

Ratio Test for Convergence of an Infinite Series

Integral Test for Convergence of an Infinite Series

Limit Comparison Test for Convergence of an Infinite Series

Alternating Series Test (Leibniz's Theorem) for Convergence of an Infinite Series

Infinite Sequences

Root Test for for Convergence of an Infinite Series

Infinite Series

Strategies to Test an Infinite Series for Convergence

Harmonic Series

Indeterminate Forms and de L'hospital's Rule

Partial Sums of Infinite Series