Tuesday, January 5, 2016

calculus - Calculate the value of the series ,sumin=1nftyfrac12n(2n+1)(2n+2)


Calculate the infinite sum



1234+1456+1678+



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(12i22i+1+12i+2)

Then ni=112i(2i+1)(2i+2)=ni=1(12i12i+1)14+12(2n+2)=ni=112i(2i+1)14+12n(2n+2)
Clearly ni=112i(2i+1)=10x0(t+t3+t5++t2n1)dtdx10x0tdt1t2dx=1210x0(11t11+t)dtdx=1210(log(1x)+log(1+x))dx=
We have 10log(1x)dx=10logxdx=xlogxx|10=1,10log(1+x)dx=21logxdx=xlogxx|21=2log21.
Hence i=112i(2i+1)=1log2
and finally i=112i(2i+1)(2i+2)=34log2.


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