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1

Fourier series expansion works because $\sin(nx),\cos(nx)$ are an orthonormal basis (after normalizing). You will need to massage things a bit, in order to turn $\sinh(nx),\cosh(nx)$ into an orthonormal basis. For instance, if you are trying to approximate a function on $[0,1]$, you will need your expansion to satisfy $$ \int_0^1 \sinh(nx)\sinh(mx)=0 $$ when ...


0

Writing $z = x + iy$, we have that $$ \left| \frac{z-1}{z+1} \right|^2 = \frac{(x-1)^2+y^2}{(x+1)^2 + y^2} < 1.$$ Thus, the series converges pointwise. You can deduce local uniform convergence by using continuity of absolute value. Fixing $z_0 \in \text{Re}{z} > 0$, choose $s$ with $0 < s < 1$ and $r > 0$ so that for all $z \in B(z_0,r)$, ...


2

What does the expression "$e^{2x} - e^{2x} + e^{2x} - e^{2x} + \cdots$" mean? It is certainly not a function under the usual sense of "limit of partial sums." They don't even converge pointwise! So the problem is that you're not being careful with infinity --- your argument hinges on using this symbol of an infinite alternating sum of $e^{2x}$s, and you ...


0

We need to show that $x_{n+1}-x_n<0$ for all $n\ge 1$ whenever x_0=a>1$ First note that if $x_n>1$, then $x_{n+1}=\sqrt{2x_n-1}>1$ also. Next, to show $x_{n+1}-x_n<0$, all we need to show is that $2x_n-1<x_n^2$. But this is trivially the same as $x_n^2-2x_n+1=(x_n-1)^2>0$. And since we know $x_n>1$, then we are done!


0

Since $$\lim_{n \to \infty} \frac{(-3)^{n-1}}{n^5} = \infty \neq 0,$$ the series diverges by the $n$th term test.


0

Hint: It is enough to show the graph of $y=\sqrt{2x-1}$ is under the straight line $y=x$. For an increasing sequence it should be over this line.


0

The condition to prove is that $x_{n+1}<x_n$, that is, $$ \sqrt{2x_n-1}<x_n $$ that's equivalent to $$ 2x_n-1<x_n^2 $$ or $$ x_n^2-2x_n+1>0 $$ that's true provided you prove in advance that $x_n>1/2$, for all $n$, in order that the expressions are meaningful. Try proving, instead, that $x_n>1$, by induction on $n$.


0

All you need to show is that the the sequence $x_{n}=(-1)^{n}$ does not converge. Many ways to do it : First, using the theorem saying that if a sequence converges , then any subsequence converges to the same limit. $x_{2n}$ and $x_{2n+1}$ are subsequences converging obviously to different limits. Not familiar with this? No problem! Assume that $x_{n}$ ...


0

One can view it as a function $a_n = f(a_{n-1})$ whereas $f(x) = \sqrt{2x-1}$, then the monotonicity of $a_n$ depends on $f'(x)$. And $f'(x) = \dfrac{1}{\sqrt{2x-1}} > 0$, this means the sequence is increasing as stated.


0

Use induction. Your base case is $a > \sqrt{2a - 1} \Leftrightarrow a^2 > 2a - 1 \Leftrightarrow (a-1)^2 > 0$, which is true since $a > 1$. Furthermore, notice that $x_1 = \sqrt{2a - 1} > 1$. Inductive step: assume that $x_k < x_{k-1}$ and $x_k > 1$. We wish to show that $x_{k+1} < x_k$. Well, we have $x_{k+1} = \sqrt{2 x_k - 1}$, ...


0

If it converges to $\ell$ then $|s_n-\ell|<\varepsilon$, once $\varepsilon>0$ is fixed, and $n$ is sufficiently large. Here $s_n$ is exactly the $n$-th partial sum. But if you set, for example, $\ell=1/10$, then the above inequality cannot be satisfied even for $n$ enough large.


1

Hint: you can show by induction that $\{x_n\}$ is monotonically increasing and bounded. This implies that the sequence converges to say $L$. Now, $$\lim_{n\to\infty}x_{n+1}=\sqrt{1+\lim_{n\to\infty}x_n}$$ implies $$L=\sqrt{1+L}.$$


2

When we say a series converges, we mean that the sequence of partial sums approaches a limit. Let's consider the sequence of partial sums of your series: $$\begin{matrix}-1 & = & -1 \\ -1 + 1 &= &0 \\ -1 + 1 - 1 &= &-1\end{matrix}$$ and so on, so it keeps alternating from $-1$ to $0$. It therefore doesn't approach a unique value. The ...


2

That depends on how you are planning to use that sum. I believe that there are physicist who will insist on the expression being a half, but mathematically there is no convergence. The partial sums are zero and one alternatingly, therefore no limit exists. If you go for the physics approach you will get the value thus: Let $S=\sum_{n=1}^\infty (-1)^n$ ...


3

Hint: If $\sum a_n$ converges, you must have $a_n \to 0$. Does this happen here?


3

You messed up your indexing ($k$ doesn't appear in your summand and the upper limit must be $\infty$), but the terms are okay. The correct representation would be $$e^x = \sum_{n=0}^\infty \frac{e^3}{n!} (x-3)^n$$ Derivation is pretty straight-forward using the known expansion at $0$ and $e^{a+b}=e^a e^b$: $$e^x = e^{3+(x-3)} = e^3 e^{x-3} = e^3 ...


9

The sum converges because $\sin(n)\leq 1$, so \begin{equation} \sum_{n=1}^{\infty} \Bigl| \frac{\sin^7(n)}{(n^7+1)^{1/2}} \Bigr| \leq \sum_{n=1}^{\infty} \frac{1}{(n^7+1)^{1/2}} < \sum_{n=1}^{\infty} \frac{1}{(n^7)^{1/2}} = \sum_{n=1}^{\infty} \frac{1}{n^{7/2}} \end{equation} The last series converges, so the given series converges absolutely. Absolute ...


5

You are not applying well the properties of the exponentials. Instead, you can write: $$ \sum_{n=1}^\infty \Bigg| \frac{\sin(n)^7}{(n^7)^{1/2}}\Bigg| \leq \sum_{n=1}^\infty \frac{1}{n^{7/2}} <\infty,$$ since $$ \sum_{n=1}^\infty \frac{1}{n^{\alpha}}<\infty $$ if and only if $\alpha>1$.


1

Exists a closed form in terms of special functions. We have $$\sum_{k\geq1}\frac{1}{ka\left(\left(ka\right)^{2}-1\right)}=\frac{1}{a^{3}}\sum_{k\geq1}\frac{1}{k\left(k^{2}-\frac{1}{a^{2}}\right)}=\frac{1}{a^{3}}\sum_{k\geq1}\frac{1}{k\left(k-\frac{1}{a}\right)\left(k+\frac{1}{a}\right)} $$ and using the identity ...


0

Assume that $f(z)$ is an even analytic function in a neighbourhood of zero, $$ f(z) = \sum_{k\geq 0} a_k z^{k}. $$ We have: $$ 0=f(z)-f(-z) = 2\,\sum_{k\geq 0} a_{2k+1}\, z^{2k+1} $$ hence for every $k\geq 0$, $a_{2k+1}=0$.


0

Comment turned answer per request. Just multiply the $n^{th}$ equation by $z^n$ and start to sum from $n = 2$. $$ \overbrace{\overset{*}{F}(z) - g(0) - g(1)z}^{ \sum\limits_{n=2}^\infty\overset{*}{F}_n z^n} = \overbrace{z(\overset{*}{F}(z)-g(0))}^{ \sum\limits_{n=2}^\infty \overset{*}{F}_{n-1} z^n } + \overbrace{z^2\overset{*}{F}(z)}^{ ...


0

For $|x| > 1$, the sequence $x^{n!}$ is not even bounded, so the series does not converge. For $x = 1$, the sequence of partial sums is not bounded ($S_n = n$), so it does not converge. For $x = -1$, the sequence has periodic behavior ($S_{2n} = 1, S_{2n+1}=0$) and does not converge either. For $|x| < 1$ the series $\sum x^n$ converges absolutely ...


1

Not a comprehensive answer yet, but still food for thought : For all $k$, there is a systematic way to compute a closed form formula for $a_k(n)$ valid for all $n$. The steps are the following : You can easily prove that for all $k$, $a_k(n)$ is a polynomial function of the variable $n$. Proof : It is true for $k=0$ since $a_O(n) = 1$. If $a_{k-1}(n)$ is ...


1

$$\left(p+q\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}p^{k}q^{n-k}$$ Taking the derivative w.r.t. $p$ on both sides we find: $$n\left(p+q\right)^{n-1}=\sum_{k=0}^{n}\binom{n}{k}kp^{k-1}q^{n-k}$$ Multiplying with $p$ on both sides we find: $$np\left(p+q\right)^{n-1}=\sum_{k=0}^{n}\binom{n}{k}kp^{k}q^{n-k}$$ If $p+q=1$ then this equality can be written as: ...


0

You are not able to see what was done. Just take out $n^2$ common from numerator and Denominator, which will then cancel out. So, you basically have $$\frac{n^2(1+\frac{1}{n^2})}{n^2(2-\frac{3}{n^2})}$$ Do you see it now?


0

Hint. Observe that, using the Mean Value Theorem, we obtain that there is a $\vartheta\in(0,1)$, such that $$ \sin\frac{1}{n}=\sin\frac{1}{n}-\sin 0=\frac{1}{n}\cos\big(\vartheta\frac{1}{n}\big). $$ Hence $$ 0< 1-n\sin\frac{1}{n}=1-\cos\big(\vartheta\frac{1}{n}\big)=2\sin^2\big(\frac{\vartheta}{2n}\big)\le ...


0

Nothing too difficult, really: $$\frac{n^2+1}{2n^2-3} = 1\cdot \frac{n^2+1}{2n^2-3} = \frac{\frac{1}{n^2}}{\frac{1}{n^2}}\cdot \frac{n^2+1}{2n^2-3} = \frac{\frac{1}{n^2}(n^2+1)}{\frac{1}{n^2}(2n^2-3)} = \frac{1 + \frac{1}{n^2}}{2 - \frac{3}{n^2}}$$


1

Another possible approach (which may be off-topic) is to use the Taylor series expansion of $$\sin(x)=\sum_{i=0}^\infty \frac {(-1)^i}{(2i+1)!}x^{2i+1}$$ and to replace $x$ by $\frac 1n$.So, $$1-n\sin(\frac 1n)=-n\sum_{i=1}^\infty \frac {(-1)^i}{(2i+1)!}\frac 1 {n^{2i+1}}=-\sum_{i=1}^\infty \frac {(-1)^i}{(2i+1)!}\frac 1 {n^{2i}}$$ Summing over $n$ from ...


1

If $f_n$ converges converges uniformly on each of finitely many sets, then it converges uniformly on the union of these sets. The proof is of the "let $N=\max (N_1, \dots, N_k)$" variety. So yes to your first question. (This doesn't really have much to do with continuity.) No, on the second question: Let $D = [0,1/2] \cup \{2/3, 3/4, 4/5, \dots \}.$ Define ...


3

There are several ways to show that the series converges. I thought it would be instructive to show an application of Gauss's Test. To that end, we note that $$\begin{align} a_n&=1- n\,\sin (1/n)\\\\ &=1-n\left(\frac{1}{n}-\frac16 \frac{1}{n^3}+O\left(\frac{1}{n^5}\right)\right)\\\\ &= \frac{1}{6n^2}+O\left(\frac{1}{n^4}\right) ...


0

Hint: First show $1 - x \sin(1/x) \to 0$ as $x \to \infty$. Then do a limit comparison test comparing the ratio of terms to $1/n^2$, using L'Hopital's rule.


4

By mean value theorem, we have $n\sin\frac 1n = \frac{\sin\frac{1}{n}}{\frac 1n} = \cos(\theta)$ for some $\theta\in [0,\frac{1}{n}]$. Hence $1 \geq n\sin\frac 1n \geq \cos\frac 1n$. It follows, that $0 \leq 1-n\sin\frac 1n\leq 1-\cos\frac{1}{n}$. By mean value theorem, we have $1-\cos\frac{1}{n} = \sin(\phi)\frac 1 n$ for some $\phi\in [0,\frac 1n]$. It ...


2

If $f(n) = O(n^\theta)$ for some $\theta \in \mathbb{R}$ then we have for every $s>0$, $\frac{f(n)}{n^{\theta+s}} \to 0$ as $n\to \infty$. In particular $$\frac{|f(n)|}{n^{\theta+s}} < \frac{C}{n^s}$$ for large enough $n$. $$\left| \sum_{n=1}^\infty \frac{f(n)}{n^{\theta+s}} \right| \le \sum_{n=1}^\infty \frac{|f(n)|}{n^{\theta+s}} \le C' ...


2

First of all, note that you cannot use the expansion $$ \frac{1}{1 - z} = 1 + z + z^2 + … \text{,}$$ because the condition $|z| < 1$ does not necessarily hold. In fact, it would be undesirable to produce any power series expansion in $z$, because you are supposed to find the Laurent series centered at $1$. Merely take your correct equation $$ ...


1

$$ a_0=0, a_1=1, a_n=\frac{1}{2}\left(a_{n-1} + a_{n-2} \right) \tag 1 $$ is a Linear homogeneous recurrence relations with constant coefficients of the order $2$. As explained in that Wikipedia article, the solution has the form $$ a_n = C \lambda_1^n + D \lambda_2^n $$ where $\lambda_{1,2}$ are solutions of the "characteristic equation", in this case $$ ...


1

If you take the coefficient of all terms to be $2^n2^m$ for sufficiently large $m$ then you will be able to successfully extract the first $n$ bits after the decimal point. $m = k_0 + 1$ will certainly work but $m = O(\log k_0)$ should also work. The point is that at most you will lose 1 bit of extra contribution at the $(n + m)$th place after the decimal ...


1

This is a geometric series. Since $$(1+x+x^2+\cdot+x^n)(1-x)=1-x^{n+1}$$ then $$s_n=\frac{1-\frac{1}{10^{n+1}}}{1-\frac1{10}}$$ and $\lim s_n=\frac{10}9$. Also, this is $$1.1111\ldots11$$ which is the ninth part of $$9.999\ldots99<10$$


0

Instead, try to find an explicit formula for the terms $s_n$. Prove that the formula holds by induction, then proceed from there. To get an idea of an explicit formula, calculate the first several terms, and look for a pattern. Added: As an alternative, try proving that $$\frac{10}9-s_n=\frac1{10^{n+1}\cdot 9}$$ for all $n$, using induction and the ...


1

hint: $s_n-s_{n-1} = \dfrac{1}{10^n}, s_{n-1}-s_{n-2} = \dfrac{1}{10^{n-1}}, s_{n-2}-s_{n-3} = \dfrac{1}{10^{n-2}},...$, and : $s_n = (s_n-s_{n-1})+(s_{n-1}-s_{n-2})+\cdots + (s_2-s_1)+(s_1-s_0)+s_0$.


0

Hint: Call $S=\sum_{k=0}^\infty \dfrac{1}{10^k}$


4

Let $\sum u_n$ the given series. It is not difficult to see that $$n^\alpha u_n\xrightarrow{n\to\infty}0\quad \forall 1<\alpha<\frac32$$ so $$\exists n_0,\quad \forall n\ge n_0,\; u_n\le \frac1{n^\alpha}$$ hence the series $\sum u_n$ is convergent by comparison with the Riemann series.


3

$$\text{Consider }S = \sum_{i=0}^N r^i$$ $$(1 - r)S = \big(1 + r + r^2 + \ ... \ + r^N\big) - \big(r + r^2 + ... + r^{N+1}\big) = 1 - r^{N+1}$$ So $$S = \frac{1 - r^{N+1}}{1 - r} = \frac{r^{N+1} - 1}{r - 1}$$ In the case of $3$, $S_3 = \frac{1}{2}(3^{N+1} - 1)$


1

Consider $(\mathbb{Z},\mathcal{P}(\mathbb{Z}))$ endowed with the counting measure $$\mu(B) := \sum_{h \in \mathbb{Z}} \delta_h(B).$$ Then for any integrable function $f \in L^1$, we have $$\int f \, d\mu = \sum_{h \in \mathbb{Z}} f(h). \tag{1}$$ Define $$f_n(h) := \begin{cases} \left(1- \frac{|h|}{n} \right) \gamma(h), & |h| < n, \\ 0, & |h| ...


0

One good option here is the logarithmic test: compute $\lim \limits _{n \to \infty} \frac {\log \frac 1 {x_n}} {\log n}$. If you get a number $<1$, the series is divergent; if you get $>1$, it is convergent; if you get $1$, you cannot draw any conclusion and you must try some other method. The test works only for series with positive terms. In your ...


2

I'm assuming you aim to determine the convergence of $$\sum_{n=2}^\infty \frac{1}{\sqrt{n}\log(n)}$$ Let's try to find a series that is smaller than this one, but still divergent. For example, $$\sum_{n=2}^\infty \frac{1}{n\log(n)}$$ By the integral test we have $$\int_{2}^\infty \frac{1}{x\log(x)}\text{d}x = \log(\log(x))\Big\vert^\infty_2 = \infty $$ so ...


0

For b) with ratio test: $$ \lim\limits_{n \rightarrow \infty}{\frac{cosh(n)}{cosh(n+1)}} = \lim\limits_{n \rightarrow \infty}{\frac{e^k+e^{-k}}{e^{k+1}+e^{-k-1}}} = \lim\limits_{n \rightarrow \infty}{\frac{1+e^{-2k}}{e+e^{-2k-1}}} =\frac{1}{e} $$


0

To do b) with the ratio test, consider that the ratio between terms is $$\frac{\cosh n}{\cosh(n+1)} = \frac{e^n+e^{-n}}{e^{n+1}+e^{-n-1}}\to e^{-1}<1$$ The hard part is rigorously showing the convergence but the idea is that the $e^{-n}$ terms become negligible as $n\to\infty$. EDIT: This isn't hard. Because $\lim_{n\to\infty}e^{-n}=0,$ ...


2

For (a) you do not need the ratio test: just note that $\lim \limits _{n \to \infty} \frac {\sinh n} {\mathbb e ^n} = \lim \limits _{n \to \infty} \frac {\mathbb e ^n - \mathbb e ^{-n}} {2 \mathbb e ^n} = \frac 1 2$. By the zero test, since the limit is not $0$ then the series diverges. For (b) indeed, one could use the ratio test, but I shall use the root ...



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