Prove that series converges uniformly

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SUMMARY

The discussion focuses on proving the uniform convergence of two series: (a) $\displaystyle\sum_{n=1}^\infty{\displaystyle\frac{ln(1+nx)}{nx^n}}$ on the set $S = [2, \infty)$ and (b) $\displaystyle\sum_{n=1}^\infty{(-1)^{n+1} \displaystyle\frac{e^{-nt}}{\sqrt[ ]{n+t^2}}}$ on the set $S = [0, \infty)$. For series (a), the Weierstrass M-test is applied, using the inequality $\ln(1+t) \le t$ to show that the $n$th term is dominated by $\dfrac{1}{2^{n-1}}$. For series (b), it is established that the terms are alternating and decreasing, which is sufficient for uniform convergence.

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fabiancillo
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I was studying uniform convergence. I have doubts

a) Prove that series $\displaystyle\sum_{n=1}^\infty{\displaystyle\frac{ln(1+nx)}{nx^n}}$
converges uniformly on the set $ S = [2, \infty) $.

b Prove that series $\displaystyle\sum_{n=1}^\infty{(-1)^{n+1} \displaystyle\frac{e^{-nt}}{\sqrt[ ]{n+t^2}}}$
converges uniformly on the set $ S = [0, \infty) $.

My attempt:

a) The fuctions $f_n(x)=\displaystyle\frac{ln(1+nx)}{nx^n}$ are decreasing (I don't how prove). Therefore $|f_n(x)|\leq f_n(2)$

b) I do not know how to start

Thanks
 
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Apply the Weierstrass M-test in series (a). Use the inequality $\ln(1+t) \le t$ for all $t \ge 0$ to show that the $n$th term is dominated by $\dfrac1{2^{n-1}}$.

For (b), note that the series is alternating. So it suffices to show that $\dfrac{e^{-nt}}{\sqrt{n+t^2}}$ is decreasing (by showing its derivative is negative) and converges uniformly to $0$.
 
Ok but I don't understand how to test this:

show that the $nth$ term is dominated by
$\dfrac1{2^{n-1}}$
 
Let $x\ge 2$. Since $\ln(1+ nx)\le nx$, then $$\frac{\ln(1+nx)}{nx^n}\le\frac{nx}{nx^n} = \frac{1}{x^{n-1}} \le \frac{1}{2^{n-1}}$$
 

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