Let $\sum_{n=1}^{\infty}n^5(\frac{x}{x+2})^n=S(x)$. Prove that the sum S(x) is a function and continuous to $x\epsilon [0,10]$

Question

Let $\sum_{n=1}^{\infty}n^5(\frac{x}{x+2})^n=S(x)$. Prove that the sum S(x) is a function and continuous to $x\epsilon [0,10]$

Since we are talking about sums and we need to prove continuous i directly thought about Uniform convergence.

But i can't see how to do it on this one, What's the idea?

Answer 1

Hint: For convergence, compare with $\displaystyle\sum_1^\infty n^5 \left(\frac{11}{12}\right)^n$. This can also be used to show uniform convergence in our inteval. (Irrelevant comment; the given series is not a power series.)

Answer 2

Ratio (quotient) test:

$$a_n:=n^5\left(\frac x{x+2}\right)^n\implies\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(n+1)^5x^{n+1}}{(x+2)^{n+1}}\cdot\frac{(x+2)^n}{n^5x^n}\right|=$$

$$=\left(\frac{n+1}n\right)^5\frac{|x|}{|x+2|}\xrightarrow[n\to\infty]{}\frac{|x|}{|x+2|}<1\iff |x|<|x+2|\implies x^2<x^2+4x+4\implies$$

$$\implies x>-1$$

So in any closed finite interval $\,[a,b]\;,\;\;a>-1\;$ we have uniform convergence