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1. Let \(a_n, b_n \in \mathbb{R} \) sequences and \( (a_n+b_n)b_n \neq 0 , \;\; \forall n \in \mathbb{N} \). If the series \( \displaystyle \sum_{n=1}^{\infty} \dfrac{a_n}{b_n} , \;\; \sum_{n=1}^{\infty} \left( \frac{a_n}{b_n} \right)^2 \) converge prove that the series: $$\sum_{n=1}^{\infty} \dfrac{a_n}{a_n+b_n}$$ also converges.


2. Let \(x_1>0 \). We define a sequence \(x_n \) as: \(x_n=-\ln (x_1+x_2+\cdots+x_n ) \) . Sum the series: \( \displaystyle \sum_{n=1}^{\infty} x_n \).


3. State whether the following is true or false, justifying your answer at the same time.

If the series \( \displaystyle \sum_{n=1}^{\infty} a_n \) converges so does the series \( \displaystyle \sum_{n=1}^{\infty} a_n^2 \).

Is the opposite true?

Tolaso J Kos wrote:3. State whether the following is true or false, justifying your answer at the same time.

If the series \( \displaystyle \sum_{n=1}^{\infty} a_n \) converges so does the series \( \displaystyle \sum_{n=1}^{\infty} a_n^2 \).

Is the opposite true?

3. It's false. By Leibnitz test, the alternating series \(\displaystyle{\sum_{n=1}^{+\infty}{\frac{(-1)^{n}}{\sqrt{n}}}}\) converges, but the series \(\displaystyle{\sum_{n=1}^{+\infty}\biggl({\frac{(-1)^{n}}{\sqrt{n}}}\biggr)^2=\sum_{n=1}^{+\infty}{\frac{1}{n}}}\) it is well known that diverges.

And the opposite is false. \(\displaystyle{\sum_{n=1}^{+\infty}{\frac{1}{n^2}}=\frac{\pi^2}{6}}\) and \(\displaystyle{\sum_{n=1}^{+\infty}{\frac{1}{n}}=+\infty\,.}\)

Grigorios Kostakos wrote:

Tolaso J Kos wrote:3. State whether the following is true or false, justifying your answer at the same time.

If the series \( \displaystyle \sum_{n=1}^{\infty} a_n \) converges so does the series \( \displaystyle \sum_{n=1}^{\infty} a_n^2 \).

Is the opposite true?

3. It's false. By Leibnitz test, the alternating series \(\displaystyle{\sum_{n=1}^{+\infty}{\frac{(-1)^{n}}{\sqrt{n}}}}\) converges, but the series \(\displaystyle{\sum_{n=1}^{+\infty}\biggl({\frac{(-1)^{n}}{\sqrt{n}}}\biggr)^2=\sum_{n=1}^{+\infty}{\frac{1}{n}}}\) it is well known that diverges.

And the opposite is false. \(\displaystyle{\sum_{n=1}^{+\infty}{\frac{1}{n^2}}=\frac{\pi^2}{6}}\) and \(\displaystyle{\sum_{n=1}^{+\infty}{\frac{1}{n}}=+\infty\,.}\)

Yep. Let's see another example. The series \( \displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\ln (n+1)} \) converges since it is an alternating one, but the series \( \displaystyle \sum_{n=1}^{\infty} \frac{1}{\ln^2 (n+1)} \)diverges from Pringsheim test.