# telescopic seriesGive an example on how to evaluate a telescoping series.

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We'll work on the following example of telescopic series:

1/(3n+1)(3n+4)

We'll write the fraction 1/(3n+1)(3n+4) as an algebraic sum of partial fractions:

1/(3n+1)(3n+4) = (1/3)[1/(3n+1) - 1/(3n + 4)]

We'll substitute n by 1 and we'll get:

s1 = 1/(3+1) - 1/(3 + 4) = (1/3)(1/4 - 1/7)

We'll substitute n by 2 and we'll get:

s2 = (1/3)(1/4 - 1/7 + 1/7 - 1/10)

We'll eliminate like terms:

s2 = (1/3)(1/4 - 1/10)

s3 = (1/3)(1/4 - 1/13)

.....................................

sn = (1/3)[1/4 - 1/(3n + 4)]

Lim sn = lim (1/3)[1/4 - 1/(3n + 4)]

lim (1/3)[1/4 - 1/(3n + 4)] = (1/3) lim [1/4 - 1/(3n + 4)], n->infinite

We'll substitute n by infinite:

(1/3) lim [1/4 - 1/(3n + 4)] = (1/3)*(1/4 - 1/infinite)

Lim sn = 1/12

So, the given sum is:

** Sum 1/(3n+1)(3n+4) = 1/12**