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I want to show that $$N\sum_{n = 2}^{\infty} \frac{t^n}{nN^n}$$ converges uniformly to $0$ as $N \to \infty$ on a bounded interval.

I have done a very few problems on showing that the power series converges uniformly on a bounded interval using Weierstrass M-test. But these problems have something in common that it just asks to show "uniform convergence" to whatever, not to a specific value such as $0$ in this problem.

I cannot find a nice definition of uniform convergence of series. But series is just a special type of function, right?

So given $\epsilon >0$, I have to show an $N_0 \in \mathbb{N}$ such that $N \geq N_0$ implies $|S_N(t) - 0| < \epsilon,$ where $$S_N(t) = N\sum_{k = 2}^{\infty} \frac{t^k}{kN^k},$$ for $N \in \mathbb{N}$.

But I cannot proceed further.

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  • $\begingroup$ What can you say about the general term of the series if you know that $t$ belongs to a bounded interval, that is, there exists some constant $C>0$ so that $|t|\leq C$? $\endgroup$ – uniquesolution Oct 21 '15 at 19:47
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If $|{t / N}| > 1$ then the series is divergent, so, wlog, assume $N > t^{-1}$. For $\enspace {t / N} < 1$, $$ \sum_{n=1}^\infty {(t/N)^n \over n} = -\ln(1-t/N),$$ so $$ N \sum_{k=2}^\infty {t^n \over nN^n} = N\left[ -\ln(1-t/N) - t/N \right]= -N\ln(1-t/N) - t$$ Apply L'Hopital's rule.

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Here's one possibility: $$ \left|N\sum_{k=2}^{\infty}\frac{t^{k}}{kN^{k}}\right| \leq N\sum_{k=2}^{\infty}\left|\frac{t^{k}}{N^2k}\right| \leq \frac{1}{N}\sum_{k=1}^{\infty}\left|\frac{t^k}{k}\right|$$ Now the power series on the right is the taylor expansion of the function $\ln(1+x)$ around $0$ (for positive t), which converges absolutely for any bounded and closed subinterval of $(-1,1]$, e.g. [-1/2,1/2]. As we know it converges to $\ln(1+x)$ which is bounded on $[-1/2,1/2]$, so there is a constant $C>0$ such that $$\sum_{k=1}^{\infty}\left|\frac{t^k}{k} \right| \leq C, \quad t \in [-1/2,1/2]$$

We conclude $$ \left|N\sum_{k=2}^{\infty}\frac{t^{k}}{kN^{k}}\right| \leq N\sum_{k=2}^{\infty}\left|\frac{t^{k}}{kN^{k}}\right| \leq \frac{C}{N}, \quad t\in [-1/2,1/2]$$ Taking the limit of $N$ to $\infty$ proves the claim.

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