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I'm trying to compute $\sum \tan(1/n)$ and in doing I used the limit comparison test to show this diverges. To this end, I compute $$\lim_{n \to \infty} \frac{\sin(1/n)}{\cos(1/n)})n=\lim_{n \to 0} \frac{\sin(n)}{n\cos(n)}$$

Finally I expand the trig functions by their Taylor series then move the limit inside the sum ( this is the step I'm afraid of) and finally I apply LHospital to get that the limit is $1$ and therefore the series diverges. I'm not really interested in alternate proofs that this series diverges. I just want to know if and why I can move the limit inside the series.

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    I answered, but on closer examination of the question, I don't really get it. For the limit comparison, you're comparing to the series $\sum_n1$ which diverges? Then, for the limit, are you moving the limit inside the taylor series, first to prove that the limits are both zero, then differentiating term-by-term, and then moving the limit inside again to compute the $1$? Basically you want to know why if you have a Taylor series $f(n)$ you can plug in $n=0$? I.e. you want to establish that the taylor series is continuous at zero? As I said, I'm not sure why you're expanding in taylor series. – 2017-02-15
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    I'm comparing it to the harmonic series. I used the fact that $x/(1/y)=xy$ – 2017-02-15

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In computing the limit there is no need to expand in Taylor series. You can use L'Hospital directly on $\frac{\sin(n)}{n\cos(n)}.$ Just take derivative of top and bottom and get $\frac{\cos(n)}{\cos(n) + n\sin(n)} \to 1$

However, if you want to justify doing your limit computations on the Taylor series, you're right to be careful. For instance if instead of $n\to 0$ it were $n\to\infty,$ and say you wanted to take $$ \lim_{n\to\infty} e^{-n}= \lim_{n\to\infty} \sum_{j=0}^\infty \frac{(-n)^{j}}{j!}.$$ Then moving the limit inside would be disasterous.

But when you're taking the limit $n\to0$ of a taylor series $f(n) = \sum a_jn^j$ it's generally perfectly okay to plug in $n=0$ when taking the limit and get $\lim_{n\to 0}f(n) = a_0.$ In other words, the Taylor series is continuous at zero.

To see this, note that all your power series in question have a nonzero radius of convergence and thus converge uniformly on some neighborhood of the origin. This immediately implies continuity of power series in this neighborhood (and thus at the origin) since it's the uniform limit of polynomials (which are continuous). This uniformness also justifies you differentiating term-by term.

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    What? I think you're being extremely cavalier, in my humble opinion. Check your limit. You put $1$... It's actually zero. – 2017-02-15
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    @Clclstdnt Ahh sorry, yes I was thrown off by not understanding your overall strategy. But in any event now it's clear to me why you want to compute that limit to be $1.$ Still, I think my question of why you're bothering to expand in taylor series instead of applying l'hospital directly on the functions is valid. However, I can tell you why the Taylor series are continuous at zero if you'd like. (which would answer your question of why you could do the limit term by term) – 2017-02-15