`sum_(n=0)^oo (1/2^n -1/3^n)` Find the sum of the convergent series.

`sum_(n=0)^oo(1/2^n-1/3^n)`

`=sum_(n=0)^oo1/2^n-sum_(n=0)^oo1/3^n`

`=(1/2^0+1/2^1+1/2^2+.......+1/2^oo)-(1/3^0+1/3^1+1/3^2+......+1/3^oo)`

`=(1+1/2+1/2^2+.......1/2^oo)-(1+1/3+1/3^2+........+1/3^oo)`

Now both of the above are geometric series,having first term 1 and common ratios `1/2,1/3` respectively,

Geometric series converges to the sum:

`sum_(n=0)^ooar^n=a/(1-r),0<|r|<1`

Using the above,

`=(1/(1-1/2))-(1/(1-1/3))`

`=(1/(1/2))-(1/(2/3))`

`=2-3/2`

`=1/2`

Comments

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