Let rth term of a series be given by Tr=r1−3r2+r4. Then −2∞∑r=1Tr is
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Solution
Tr=r1−3r2+r4=r(r2−1)2−r2=12[1r2−1−r−1r2−1+r]
So, n∑r=1Tr =12[−1−1]+12[1−15]+12[15−111]+12[111−119]+⋮⋮+12[1n2−1−n−1n2−1+n]Sn=12[−1−1n2−1+n]
For very large value of denominator, the fraction will be ≅0 ∴S∞=−12