What is the value of- 13!+25!+37!+49!+⋯ I wrote it as general term ∑n(2n+1)!. As the series converges it should be telescopic (my thought). But i dont know how to proceed. I also know ∑1n!=e Any help /hints appreciated. Thanks!
Answer
One may write
∞∑n=0n(2n+1)!=12∞∑n=0(2n+1)−1(2n+1)!=12∞∑n=01(2n)!−12∞∑n=01(2n+1)!=12(e+e−12)−12(e−e−12)=12⋅e−1
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