Htykuut

I'm working on the following problem:

I've observed that division of zero will occur in a single term of the series if a natural number is selected, which is why that constraint is given.

I've computed a few terms manually for x=1.1, and I can see that the terms' absolute values get smaller quite rapidly.

So, it does appear that the series will converge. My tools to prove this thus far are the M-test and the Cauchy Condition, but I'm unsure if either is suitable in this case. I was also thinking about finding another series of functions that also converges, but has larger terms than the one in this series of functions.

Any assistance would be appreciated. Thank you!

A variant of a preceding answer, since it's almost impossible to do something extremely different:

For $nin Bbb N$ let $g_n=sum_{j=n}^{infty}1/j^2.$ We have $lim_{nto infty}g_n=0.$

Let $cin Bbb N$ with $cgeqmax (|a|,|b|).$ For $c<jin Bbb N$ and for any $xin [a,b]$ we have $$|j(x-j)|=j|j-x|geq j(|j|-|x|)=j(j-|x|)geq j(j-c)geq (j-c)^2.$$

So if $c<nin Bbb N$ then $$sup_{nleq n_1leq n_2}|sum_{j=n_1}^{n_2}1/j(x-j)|leq sum_{j=n}^{infty}|1/j(x-j)|leq$$ $$leq sum_{j=n}^{infty}1/(j-c)^2=g_{n-c}.$$

And $g_{n-c}to 0$ as $nto infty.$ So the series represented by $f$ satisfies the Cauchy Condition for convergence at each $xin [a,b]$ so it converges to some $f(x).$

And for $n>c$ we have $$lim_{c<nto infty}sup_{xin [a,b]}|f(x)-sum_{j=1}^n 1/j(x-j)|leq lim_{c<nto infty}g_{1+n-c}=0$$ so the convergence is uniform.

You can build a series $u_n$ which is always bigger than $|frac{1}{n(x-n)}|$

We know that $exists ninBbb{N}, n<a<b<n+1$ because there are no natural numbers in $[a,b]$

Thus, $forall n in Bbb{N^*}, forall xin[a,b], |frac{1}{n(x-n)}|le {begin{matrix}frac{1}{n(a-n)} if n<a \ frac{1}{n(n-b)} if n>b end{matrix} = u_n$

The series we just created always converges because the series is always positive and $u_nsimfrac{1}{n^2}$

$sim$ means "equivalent". It's an asymptotic analysis of a series or function Wiki article here