(b) $\dis \frac{3n^3-1}{2n^3+1} = \frac{3-\frac{1}{n^3}}{2+\frac{1}{n^3}}\to \frac{3}{2}$.
(c) $n^{2/n}=(n^{1/n})^2\to 1^2=1$. Here we use the well-known limit \[
\limn \sqrt[n]{n}=\limn n^{1/n}=1.
\] This limit is not hard to prove. Consider $y=x^{1/x}$ for $x>0$, then $\ln y = \frac{\ln x}{x}$, by l'hopital's rule, \[
\lim_{x\to\infty} \ln y = \lim_{x\to\infty} \frac{1/x}{1} = 0,
\] so $\lim_{x\to\infty} y = \lim_{x\to\infty} x^{1/x}=e^0=1$. Since the value of $x^{1/x}$ tends to be 1 when $x$ is large, in particular, if we restrict to integer, we still have $n^{1/n}\to 1$ when $n\to \infty$.
(d) Since \[
\brac{\frac{n}{n+5}}^n =\frac{1}{(1+\frac{5}{n})^n} =\frac{1}{\big((1+\frac{1}{\frac{n}{5}})^{\frac{n}{\color{blue}5}\big)^{\color{blue} 5}}}\to \frac{1}{e^5}.
\] Here we used the limit \[
\lim_{x\to\infty} \brac{1+\frac{1}{x}}^x = e,
\] which can be derived from l'hopital's rule.
(e) $\dis \frac{\ln(1/n)}{n} = \ln \brac{\frac{1}{n}}^{1/n} = \ln \frac{1}{n^{1/n}} \to \ln \frac{1}{1}=0$.
(f) By definition of limit of sequence, only the behaviour of $\{b_n\}$ when $n$ is large is of interest. Hence \[
\limn b_n = \limn \frac{n}{e^n}=0,
\] again the last limit follows from $\lim_{x\to\infty} \frac{x}{e^x}=0$, which can be derived from l'hopital's rule.
(g) Unlike previous parts the sequence here is divergent. Let's recall the following result:
In particular, if we can show there are two subsequences of $\{x_n\}$ converging to two different limits, then we conclude $\{x_n\}$ diverges.A sequence $\{x_n\}$ converges to $\ell$ if and only if every subsequence of $\{x_n\}$ converges to $\ell$.
The subsequence $\{(-1)^{2n}\frac{2n}{2n+1}\}$ converges to 1
The subsequence $\{(-1)^{2n-1}\frac{2n-1}{2n-1+1}\}$ converges to $-1$
Conclusion: $\{(-1)^n \frac{n}{n+1}\}$ diverges.
Problem 2. To argue a sequence converges to 0, we use the following standard method:
The reason is simple, since $-|a_n|\leq a_n\leq |a_n|$ is always true, hence if $|a_n|\to 0$, then $a_n\to 0$ by Sandwich theorem.If $|a_n|\to 0$, then $a_n\to 0$.
Now we compute the limit:
(a) Since $|\frac{\cos n}{n}|\leq \frac{1}{n}$, thus $|\frac{\cos n}{n}|\to 0$ by Sandwich theorem, so $\frac{\cos n}{n}\to 0$.
(b) Since $|\frac{\sin 6n}{5n}|\leq \frac{1}{5n}$, so $|\frac{\sin 6n}{5n}|\to 0$ by Sandwich theorem, so $\frac{\sin 6n}{5n}\to 0$.
(c) Since $|\frac{2\tan^{-1}n}{n^3+4}|\leq \frac{2(\pi /2)}{n^3+4}$, so $|\frac{2\tan^{-1}n}{n^3+4}|\to 0$ by Sandwich theorem, and thus $\frac{2\tan^{-1}n}{n^3+4}\to 0$.
(d) Method 1. Since \[
\brac{1-\frac{1}{2}}\brac{1-\frac{1}{3}}\brac{1-\frac{1}{4}}\cdots \brac{1-\frac{1}{n}} =\frac{1}{\color{blue}2} \frac{\color{blue}2}{\color{green} 3} \frac{\color{green} 3}{4}\cdots \frac{n-1}{n}=\frac{1}{n},
\] hence $\brac{1-\frac{1}{2}}\brac{1-\frac{1}{3}}\cdots \brac{1-\frac{1}{n}}\to 0$ obviously from this formula.
Method 2. Recall that $1+x\leq e^x$ on $\R$, this can be easily proved. In fact this statement has a very clear geometric intuition since the graph of $y=f(x)=e^x$ is convex while \[y = f'(0)(x-0)+f(0) = x + 1\] is the tangent line of $y=e^x$ at the point $(0,1)$.
Now we use this inequality to solve the problem, by putting $x=-1/k$ above, we have \[
1-\frac{1}{k}\leq e^{-\frac{1}{k}},
\] hence \[
\brac{1-\frac{1}{2}}\brac{1-\frac{1}{3}}\cdots \brac{1-\frac{1}{n}}\leq e^{-\frac{1}{2}}e^{-\frac{1}{3}}\cdots e^{-\frac{1}{n}}=\frac{1}{e^{\sum_{k=2}^n1/k}}.
\] Recall that the harmonic series $\sum_{k=1}^n\frac{1}{k}\to \infty$, hence by Sandwich theorem, \[\brac{1-\frac{1}{2}}\brac{1-\frac{1}{3}}\cdots \brac{1-\frac{1}{n}}\to 0.\]
Remark. Method 2 can be generalized to prove a wider range of limit. Let $a_1,a_2,\dots \in [0,1)$, then by the same method above we can show that \[
\sum_{n=1}^\infty a_n=\infty\implies (1-a_1)(1-a_2)\cdots (1-a_n)\to 0.
\] In fact $\Longleftarrow$ is also true, this requires further analysis (which is too messy to mention here).
Problem 3. (a) Let $P(n)$ be the statement: $(1+x)^n\ge 1+nx$, for all $x >-1$.
Base Case: When $n=1$, LHS $=1+x$, RHS $=1+x$, so $P(1)$ is true.
Inductive Step: Assume $P(k)$ is true for some $k$, i.e., $(1+x)^k\ge 1+kx$, for all $x>-1$, then\[
\begin{align}
\nonumber (1+x)^{k+1}&=(1+x)^k(1+x)\\
\label{1+x >= 0}&\ge (1+kx) (1+x) \\
\nonumber &=1+x+kx +kx^2\\
\label{square}&\ge 1+(k+1)x.
\end{align}
\] (\ref{1+x >= 0}) holds because $x>-1$, so $1+x>0$. Also (\ref{square}) holds since $kx^2\ge 0$, i.e., a square of a number is always bigger than or equal to zero. Now $P(k+1)$ is true. We conclude that $P(n)$ is true for each $n$.
(b) Now \[
\begin{align}
\nonumber\frac{x_{n+1}}{x_n}&=\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^n}\\
\nonumber&=\brac{\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}}^{n+1}\brac{1+\frac{1}{n}}\\
\label{we start here}&= \brac{1-\frac{1}{(n+1)^2}}^{n+1} \brac{1+\frac{1}{n}}.
\end{align}
\] Now we apply part (a) to conclude \[
\frac{x_{n+1}}{x_n}\ge \brac{1+(n+1)\brac{-\frac{1}{(n+1)^2}}}\frac{n+1}{n}=\brac{1-\frac{1}{n+1}}\frac{n+1}{n}=1.
\] Hence $x_{n+1}\ge x_n$, thus $\{x_n\}$ is increasing.
(c) Finally to show $\{x_n\}$ is bounded above, we use the binomial expansion: \[\begin{align*}
\brac{1+\frac{1}{n}}^n &=\sum_{k=0}^n \binom{n}{k}\frac{1}{n^k}\\
& =\sum_{k=0}^n \frac{n(n-1)(n-2)\cdots (n-k+1)}{k!n^k}\\
&=\sum_{k=0}^n \frac{1}{k!}\brac{1-\frac{1}{n}}\brac{1-\frac{2}{n}}\cdots \brac{1-\frac{k-1}{n}}\\
&\leq \sum_{k=0}^n \frac{1}{k!} \\
&=2+\sum_{k=2}^n \frac{1}{k!}\\
&\leq 2+\sum_{k=2}^n \frac{1}{k(k-1)} =2+\sum_{k=2}^n \brac{\frac{1}{k-1}-\frac{1}{k}} =3-\frac{1}{n}\leq 3.
\end{align*}
\] Hence by (b) and (c), $\{x_n\}$ converges.
Problem 4. Forget about this.
Problem 5.
To compute the series in (a) and (b) we just need to consider the partial sum, and then take limit.Definition. For a sequence $a_1,a_2,\dots$, the notation $\sum_{k=1}^\infty a_k$ is called infinite series, or just series, which means $\limn \sum_{k=1}^n a_k$. Here $\sum_{k=1}^n a_k$ is called the $n$th partial sum of $\sum_{k=1}^\infty a_k$ (just a name! not very important).
(a) \[
\sum_{k=0}^n \frac{1}{(3k+1)(3k+4)}=\frac{1}{3}\sum_{k=0}^n\brac{\frac{1}{3k+1} -\frac{1}{3k+4}}=\frac{1}{3}\brac{1-\frac{1}{3n+4}},
\] hence \[\sum_{k=0}^\infty \frac{1}{(3k+1)(3k+4)}=\limn \sum_{k=0}^n \frac{1}{(3k+1)(3k+4)} = \limn \frac{1}{3}\brac{1-\frac{1}{3n+4}}=\frac{1}{3}.\]
(b) $\dis \sum_{k=0}^\infty \brac{\frac{1}{4}} ^k 5^{6-k}=5^6\sum_{k=0}^\infty \brac{\frac{1}{20}}^k =\frac{20\cdot 5^6}{19}$.
(c) Since $\dis \frac{(k+1)^2}{2^{k+1}} -\frac{k^2}{2^k} =-\frac{1}{2}\frac{k^2}{2^k} +\frac{k}{2^k} +\frac{1}{2^{k+1}}$, if we take $\sum_{k=1}^n$ on both sides, we have\begin{equation}\label{finallllly}
\frac{(n+1)^2}{2^{n+1}}-\frac{1}{2}=-\frac{1}{2}\sum_{k=1}^n \frac{k^2}{2^k} +\boxed{\dis \sum_{k=1}^n \frac{k}{2^k} }+\sum_{k=1}^n \frac{1}{2^{k+1}}.
\end{equation} So the evaluate of $\sum_{k=1}^n \frac{k^2}{2^k}$ reduces down to the evaluation of the boxed term. What happens if we repeat the same process with $ \frac{k}{2^k}$ in place of $ \frac{k^2}{2^k}$? Well, we have \begin{align*}
\frac{k+1}{2^{k+1}}-\frac{k}{2^k}=-\frac{1}{2}\frac{k}{2^k}+\frac{1}{2^{k+1}}&\implies \frac{n+1}{2^{n+1}}-\frac{1}{2} =\frac{-1}{2}\sum_{k=1}^n \frac{k}{2^k}+\sum_{k=1}^n \frac{1}{2^{k+1}}\\
&\implies \sum_{k=1}^n\frac{k}{2^k}=2-\frac{n+2}{2^n}.
\end{align*} Finally, we plug in this formula in to (\ref{finallllly}) to get desired result.
Remark. By taking $n\to \infty$, we have $\dis \sum_{k=1}^\infty \frac{k^2}{2^k}=6$. Later we will see that infinite series of the form \[
\sum_{k=1}^\infty \frac{k^p}{a^k},
\] where $a>1$ and $p\in \N$ (positive integers) are very easy to compute by the technique of power series.
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