Find the sum of the series $1/2-1/3+1/4-1/9+1/8-1/27+........$ to infinity?

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Answer 1 · 2017-01-16T16:13:10

$1/2-1/3+1/4-1/9+1/8-1/27+........=1/2$

$1/2-1/3+1/4-1/9+1/8-1/27+........$ can be rewritten as

$(1/2+1/4+1/8+........)-(1/3+1/9+1/27+........)$

Hence, the given infinitive is one infinitive geometric series subtracted from another infinitive geometric series.

Observe that in the infinite series $1/2+1/4+1/8+........$ , while first term $1/2$ and common ratio is $1/2$

and in the infinite series $1/3+1/9+1/27+........$ , first term $1/3$ and common ratio is $1/3$

As for an infinite series whose first term is $a$ and common ratio $r$ is such that $|r|<1$ , the sum converges to $a/(1-r)$ .

Hence $1/2+1/4+1/8+........=(1/2)/(1-1/2)=(1/2)/(1/2)=1$ , and

$1/3+1/9+1/27+........=(1/3)/(1-1/3)=(1/3)/(2/3)=1/2$

Hence $1/2-1/3+1/4-1/9+1/8-1/27+........=1-1/2=1/2$

Find the sum of the series 1/2-1/3+1/4-1/9+1/8-1/27+........1/2-1/3+1/4-1/9+1/8-1/27+........ to infinity?