1. ## Series

For each of the series below select the letter from A to C that best applies and the letter from D to K that best applies. A possible answer is AF , for example.

A. The series is absolutely convergent.
B. The series converges, but not absolutely.
C. The series diverges.
D. The alternating series test shows the series converges.
E. The series is a -series.
F. The series is a geometric series.
G. We can decide whether this series converges by comparison with a -series.
H. We can decide whether this series converges by comparison with a geometric series.
I. Partial sums of the series telescope.
J. The terms of the series do not have limit zero.
K. None of the above reasons applies to the convergence or divergence of the series.
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2. Originally Posted by amiv4
For each of the series below select the letter from A to C that best applies and the letter from D to K that best applies. A possible answer is AF , for example.

A. The series is absolutely convergent.
B. The series converges, but not absolutely.
C. The series diverges.
D. The alternating series test shows the series converges.
E. The series is a -series.
F. The series is a geometric series.
G. We can decide whether this series converges by comparison with a -series.
H. We can decide whether this series converges by comparison with a geometric series.
I. Partial sums of the series telescope.
J. The terms of the series do not have limit zero.
K. None of the above reasons applies to the convergence or divergence of the series.
1. Ratio test
2. P-series
3. Alternating series test
4. This one seems out of place are you sure thats right
5. Harmonic series
6. Integral test
...^

3. $\displaystyle 6+\sin(n)\geq{5}$ which is how you do number 4. But if your interested

Ok, I am going to show thatt $\displaystyle \sum_{n=1}^{\infty}\frac{\sin(n)}{\sqrt{n}}$ converges
First note that

\displaystyle \begin{aligned}\sum_{n=1}^{N}\sin(n)&=\text{Im}\bi gg[\sum_{n=1}^{N}\left(e^i\right)^n\bigg]\\ &=\text{Im}\bigg[\frac{1-\left(e^i\right)^{N}}{1-e^i}\cdot{e^i}\bigg]\\ &=\frac{\cos(N)\sin(1)+\sin(N)\left(\cos(1)-1\right)-\sin(1)}{2(\cos(1)-1)}\\ &=a_N \end{aligned}

$\displaystyle |a_N|\leq{\frac{1}{2}}~\forall{N}$

So now since $\displaystyle \frac{1}{\sqrt{n}}\geq\frac{1}{\sqrt{n+1}}$ and $\displaystyle \lim_{n\to\infty}\frac{1}{\sqrt{n}}$ we can conclude that

$\displaystyle \sum_{n=1}^{\infty}\frac{\sin(n)}{\sqrt{n}}$ converges

By Dirichlet's test

4. k I got 2 as AE and 3 as BD but I'm having trouble with the other ones