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Show the convergence of the following sequences. $$a)\hspace{2mm}\left\{\frac{1}{2^{n}}+i\cdot\frac{e^{-n}}{n}\right\}\hspace{5mm}b)\hspace{2mm}\left\{1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\cdots\right\}$$ Inglés

I stalled during the demonstration of the first; I took the limit of the sequence and 0 comes out, and to show that this is the limit I used the definition,

\begin{align*} \lim_{n\rightarrow0}a_{n}&=0&\Longleftrightarrow&\forall \epsilon>0,\exists N>0&:\left|\left|\frac{1}{2^{n}}+\frac{ie^{-n}}{n}\right|\right|&<\epsilon\\& &&&:\sqrt{\frac{1}{2^{2n}}+\frac{e^{-2n}}{n^{2}}} &<\epsilon\\&&&&\leq\left|\frac{1}{2^{n}}\right|+\left|\frac{e^{-n}}{n}\right|&<\epsilon \end{align*} and to demonstrate the convergence of these series, what criteria can I use?

You can recommend bibliography.

Santiago Seeker
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    For $b)$ note that this is a geometric progression, i.e. $a_{n+1} = \frac12 a_n$. So $a_n = \left(\frac12\right)^n$, and hence for any $\varepsilon>0$ we can choose $n$ such that $n>\frac{\log\varepsilon}{-\log 2}$, from which $a_n<\varepsilon$ and we conclude that $a_n\to0$. – Math1000 Oct 21 '19 at 19:31

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For any natural $ n $ ,

$$2^n\ge n$$ and $$e^{-n} \le 1$$

thus $$|a_n|=$$ $$|\frac{1}{2^n}+i\frac{e^{-n}}{n}|\le\frac 2n$$

then, for $ \epsilon>0 $ given, to realise $|a_n|<\epsilon $ for $ n$ greater than a certain $N$, it is sufficient to have

$$\frac 2n<\epsilon \text{ or } n>\frac{2}{\epsilon}$$

So, we can take $N =\lfloor \frac{2}{\epsilon}\rfloor +1$.

This proves that the SEQUENCE $(a_n) \to 0$.

For the SERIES, use the fact that $\sum a_n$ is a sum of two convergent series. the first is geometric $\frac{1}{2^n}$ and the modulus of the second is smaller then $ e^{-n} $ which is also a convergent geometric series.

hamam_Abdallah
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