Calculate the infinite sum
12⋅3⋅4+14⋅5⋅6+16⋅7⋅8+⋯
I know this series is convergent by Comparison Test, but I can't understand how can I get the value of the sum.
Is there any easy way to calculate this?
Please someone help.
Answer
Hint. First observe that 12i(2i+1)(2i+2)=12(12i(2i+1)−1(2i+1)(2i+2))=12(12i−22i+1+12i+2)
Then n∑i=112i(2i+1)(2i+2)=n∑i=1(12i−12i+1)−14+12(2n+2)=n∑i=112i(2i+1)−14+12n(2n+2)
Clearly n∑i=112i(2i+1)=∫10∫x0(t+t3+t5+⋯+t2n−1)dtdx⟶∫10∫x0tdt1−t2dx=12∫10∫x0(11−t−11+t)dtdx=−12∫10(log(1−x)+log(1+x))dx=⋯
We have ∫10log(1−x)dx=∫10logxdx=xlogx−x|10=−1,∫10log(1+x)dx=∫21logxdx=xlogx−x|21=2log2−1.
Hence ∞∑i=112i(2i+1)=1−log2
and finally ∞∑i=112i(2i+1)(2i+2)=34−log2.
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