### Week 5 1. ... 2. ... 3. Given an increasing sequence $(a_{n})$. Suppose that it has a subsequence $(a_{n_{j}})$ that converges to $l \in \mathbb{R}$ as $j\to \infty$. Then $\forall \epsilon > 0$, $\exists J \in \mathbb{N}$ such that $\begin{align} |a_{n_{j}}-l|< \epsilon & \quad \forall j \geq J \\ \text{so } -\epsilon < a_{n_{j}} - l < \epsilon \end{align}$If $n\geq n_{J}$, then since $(a_{n})$ is increasing, $a_{n} \geq a_{n_{J}} \implies a_{n}-l\geq a_{n_{J}}-l > -\epsilon$. Choose $n_{j'} > n$, then $a_{n} < a_{n_{j'}} \implies a_{n} - l \leq a_{n_{j'}} - l< \epsilon$. So $|a_{n} -l| < \epsilon \quad \forall n \geq n_{J} $which shows that $a_{n} \to l$ as $n \to \infty$. 1. ... 2. ... 3. ... 4. ... ### Week 6 8. ... 9. ... 10. (a) For all $n \in \mathbb{N}$, $a_{n} = \frac{\sqrt{ n+1 }}{n^{2}+1} \geq 0$ since its numerator and denominator are positive. Now for $n \geq 2$, $n^{2} +2 > n^{2} -1 > 0$ so $\begin{align} a_{n} &= \frac{\sqrt{ n+1 }}{n^{2} +2} < \frac{\sqrt{ n+1 }}{n^{2}-1} = \frac{\sqrt{n+1 }}{ (n-1)(n+1)} = \frac{1}{(n-1)(n+1)^{\frac{1}{2}} } \\ &< \frac{1}{(n-1)(n-1)^{\frac{1}{2}}} = \frac{1}{(n-1)^{\frac{3}{2}}} \end{align}$ Since $\sum (n-1)^{-\frac{3}{2}}$ converges, by comparison test, $\sum a_{n}$ also converges. (b) For all $n \in \mathbb{N}$, $a_{n} = \frac{n+1}{\sqrt{ n^3 +2}}$ is positive its numerator and denominator are positive. Also $3n^{2} + 3n +1 \geq 7 > 2$ so, $a_{n} = \frac{n+1}{ \sqrt{ n^{3} +2 }} \geq \frac{n+1}{\sqrt{ n^{3} + 3n^{2}+ 3n +1 }} = \frac{n+1}{(n+1)^{\frac{3}{2}}} = \frac{1}{(n+1)^{\frac{1}{2}} }$Since $\sum (n+1)^{-\frac{1}{2}}$ diverges, by comparison test, $\sum a_{n}$ also diverges. 11. ... 12. ... ### Week 7 13. ... 14. ... 15. Consider the function $f(x) = \frac{1}{x(\log(x))^{\alpha}}$ For all $x \geq 2$, $f(x)$ is clearly non-negative. Also $f(x)$ is decreasing since $x(\log x)^{\alpha}$ is increasing. Consider the integral $I = \int_{2}^{n} x^{-1} (\log x)^{-\alpha} \, dx$Let $u = \log x$, so $dx = x \, du$. Hence $I= \int_{2}^{\log n} u^{-\alpha} \, du $So for $\alpha \neq 1$, $I= \left[ \frac{u^{-\alpha+1}}{1-\alpha} \right]_{2}^{\log n} = \frac{1}{1-\alpha} [(\log n)^{1-\alpha} -(\log 2)^{1-\alpha} ]$If $\alpha > 1$, then this is unbounded as $\log n$ is unbounded, so the sum diverges. If $\alpha < 1$, then this is bounded by $0$, so the sum converges. Now for $\alpha = 1$, $I = \int_{2}^{\log n} \frac{1}{u} \, dx = [\log u]_{2}^{\log n} = \log \log n - \log 2 $which is unbounded, so the sum diverges. 16. (a) $\lim_{ n \to \infty } \frac{1}{2n-1} = \lim_{ n \to \infty } \frac{\frac{1}{n}}{2-\frac{1}{n}} = 0$ using quotient rule and sum rule for limits (since $\lim_{ n \to \infty } 2-\frac{1}{n} = 2 \neq 0$). Also $\frac{1}{2n-1}$ is positive since $2n-1 \geq 1$. So using alternating series test $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{2n-1} < \infty$For $n \geq 2$, $\frac{1}{2n-1} > \frac{1}{n}$ so, by comparison test, the sum does not converge absolutely since $\sum \frac{1}{n}$ diverges and $\frac{1}{2n-1}$ is positive. (b) $\frac{1}{n^{2}} \to 0$ as $n \to \infty$. Also $\frac{1}{n^{2}}$ is positive so, by alternating series test, $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n^{2}} < \infty$ Now $\begin{align}\sum (-1)^{n+1} \frac{1}{n^{2}} &= \sum \frac{1}{n^{2}} - 2 \sum \frac{1}{(2n)^{2}} = \frac{\pi^{2}}{6} - \frac{1}{2} \frac{\pi^{2}}{6} = \frac{\pi^{2}}{12} \end{align}$ 18. 19. ... 20. ... 21. ... 22. ...