1/3+2/9+3/27+4/81+5/243+6/729...

Question

I have to compute this, it is an infinite series with a finite value, the problem says to set them equal to x, I think the answer is less than or equal to 1

Answer 1

Let x be the sum of the series, then x/3=1/9+2/27+3/81+...

So x-x/3=1/3+2/9+3/27+...+n/3ⁿ...-(1/9+2/27+3/81+...+n/3ⁿ⁺¹...)

Therefore 2x/3=1/3+1/9+1/27+1/81+...

Let S=sum the geometric series on the right to n terms. The common ratio is 1/3.

Therefore S/3=1/9+1/27+1/81+1/243+ and

S-S/3=2S/3=1/3 because all but the first term cancel out and the last term of the infinite series approaches zero, making S=3/2×1/3=1/2.

Therefore 2x/3=1/2, and x=3/2×1/2=3/4, so the sum of the given series is 3/4.

 

Answer 2

3/4 is the answer