Evaluate
∞∑n=0(−1)n(2n+1)3n
I can show very easily that this series converges using the alternating series test. By setting
bn=1(2n+1)3n ⇒bn≤bn+1
and
limn→∞1(2n+1)3n=0
However, what is the sum of the series? I can't find it. I tried to write it out term by term but I don't see any pattern.
Answer
Note that the power series of arctan(x) is
arctan(x)=∞∑n=0(−1)nx2n+12n+1
Also note that arctan(1√3)=π6
Therefore,
π6=∞∑n=0(−1)n(1√3)2n+12n+1=∞∑n=06√3(−1)n(2n+1)3n
Thus,
∞∑n=0(−1)n(2n+1)3n=√3π6
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