The problem is set as Summation n=0 to infinity of 3/(n^2 -2n)
I rearranged it to look like summation n=0 to infinity of (-3/2)/n + (3/2)/(n+2)
what would i do next to find the sum of the series?
I rearranged it to look like summation n=0 to infinity of (-3/2)/n + (3/2)/(n+2)
what would i do next to find the sum of the series?
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OK, so the terms of your series are -3/2 times the terms of s = ∑(n =1, ∞) 1/n - 1/(n + 2)
For this series we have
s = 1 - 1/3 + 1/2 - 1/4 + 1/3 - 1/5 + 1/4 - 1/6.....
So, with exception of 1 and 1/2, each parcel of a_n = 1/n - 1/(n + 2) ends up canceled by one of the parcels of a term with higher order. It follows that s = 1 + 1/2 = 3/2. And you series, therefore, converges to -3/2 * 3/2 = =9/4.
For this series we have
s = 1 - 1/3 + 1/2 - 1/4 + 1/3 - 1/5 + 1/4 - 1/6.....
So, with exception of 1 and 1/2, each parcel of a_n = 1/n - 1/(n + 2) ends up canceled by one of the parcels of a term with higher order. It follows that s = 1 + 1/2 = 3/2. And you series, therefore, converges to -3/2 * 3/2 = =9/4.