Question

how to find sum ? this is a question fron infinite series

Answer 1

You can see that the denominator is increasing. Hence you can say that series is convergent.

You can see that the denominators are products of two consecutive terms of the A.P.
2 5 8 11 14 and so on

The general term of this A.P is
= 2 + 3(n-1)
= 3n - 1

Now if a term is (3n - 1) it's next term is
= (3n - 1 + 3)
= (3n + 2)

So we can say that the general term of the above given series is

T(n) =

$\frac{1}{(3n - 1)(3n + 2)}$

Now this can be rearranged as

$\frac{1}{3} ( \frac{1}{(3n - 1)} - \frac{1}{(3n + 2)} )$

Now we write the series in this form.

$\frac{1}{3} ( (\frac{1}{2} - \frac{1}{5} ) + (\frac{1}{5} - \frac{1}{8} ) + ( \frac{1}{8} - \frac{1}{11} )...( \frac{1}{3n - 1} - \frac{1}{3n + 2} ) )$

Now you can see that all except the first and last term get cancelled out. Leaving us with :

$\frac{1}{3} ( \frac{1}{2} - \frac{1}{3n + 2} )$

Where n tends to ∞
Thus

$\frac{1}{3n + 2} = 0$

Since (1/∞) = 0
So we are left with

$\frac{1}{3} \times \frac{1}{2}$

Which is 1/6

Hence the sum of the given series is 1/6.

Answer 2

Answer:

Step-by-step explanation:

tn=1/(3r-1)(3r+2)

if limit exists it is convergent

if you see lim x tends infinity 1/(3r-1)(3r+2)

it is 0

therefore it is convergent series

tr=1/(3r-1)(3r+2)

tr=1/3{1/3r-1 - 1/3r+2}

put r=1 in vn-1 and n in vn

sum of n terms = 1/3{1/2-1/3n+2}

sum of infinite terms  = 1/6