My Work
I felt the best way to go about this problem was to compare it to a well known MacLaurin series. I noticed it resembled the reciprocal of the absolute value of the MacLaurin series of ln(1+x) where x=12 but I had the problem of the n in the numerator still. None of the well known series have an n in the numerator.
The Well Known MacLaurin Series
My Question
Can someone give me a hint as to which well known MacLaurin series this resembles? Once I have that I know the solution!
Answer
If you want to directly proceed via MacLaurin series, try the MacLaurin series of x2(1−x)2, which unfortunately is not there in your list. Else, proceed as follows:
Let Sm=m∑n=1n2n+1.
Sm=14+28+316+432+⋯+m−12m+m2m+1♠Sm2=18+216+332+⋯+m−22m+m−12m+1+m2m+2♢
♠−♢ now gives us
Sm2=14+18+⋯+12m+1−m2m+2
I trust you can conclude from this making use of the first MacLaurin series.
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