Evaluating sums and integrals using Taylor’s Theorem

Taylor’s theorem states that

f(x)nk=0f(k)(a)k!(xa)k=xaf(n+1)(t)n!(xt)ndt

We can use this to evaluate integrals. For example, consider f(x)=b!xb+n+1(b+n+1)!. This has f(k)(0)=0 for kn and f(n+1)(x)=xb. Hence, by Taylor’s theorem,

f(1)nk=0f(k)(0)k!=10f(n+1)(t)n!(1t)ndt
b!(b+n+1)!=1n!10tb(1t)ndt
β(b+1,n+1)=10tb(1t)ndt=b!n!(b+n+1)!

So we have determined an explicit expression for the Beta function very efficiently using Taylor’s theorem.

Taylor’s theorem could also be used to evaluate partial sums using knowledge of the infinite sum; for example, considering f(x)=k=0kxk=x(1x)2=1(1x)211x for |x|<1, and noting that f(n+1)(x)=(n+2)!(1x)(n+3)(n+1)!(1x)(n+2), we can evaluate its partial sum as

nk=0kxk=x(1x)2(n+2)(n+1)x0(1t)(n+3)(xt)ndt (n+1)x0(1t)(n+2)(xt)ndt=x(1xn)(1x)2nxn+11x
after evaluating the integrals and simplifying. In fact, this holds for all x except x=1, and if a limit towards 1 is taken, we correctly obtain n(n+1)2.
What other sort of sums and integrals can be evaluated using Taylor's Theorem? Is this an effective method for summing or integrating? Are there certain limitations we should be aware of with this method?

Answer

For what concerns the first question, it would be a nice exercise trying to evaluate the following integral
10xxdx.
For what concerns the second question, there are different necessary and sufficient conditions actually i.e. If you are integrating over an interval, the uniform convergence of Taylor series is sufficient to expand the function and interchange integral with series.

Attribution
Source : Link , Question Author : Pauly B , Answer Author : Gauge_name

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