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## New answers tagged sequences-and-series

2

$${1\over(1-x)^2}={1\over 1-x}\cdot{1\over 1-x}=\sum_{j\geq0} x^j\cdot\sum_{k\geq0}x^k =\sum_{r\geq0} x^r\left(\sum_{j+k=r}1\right)=\sum_{r\geq0}(r+1)x^r\ .$$

1

Let $S=1+2x+3x^2+4x^3+\dotsb$. \begin{align} \phantom{-x^2}S&=1+2x+3x^2+4x^3+\dotsb\\ \phantom{^2}-xS&=\phantom1-\phantom2x-2x^2-3x^3-\dotsb\\ \phantom{^2}-xS&=\phantom1-\phantom2x-2x^2-3x^3-\dotsb\\ \phantom{-}x^2S&=\phantom{1+2x+2}x^2+2x^3+\dotsb \end{align} Adding them together: \begin{align} (1-2x&+x^2)S\\ ...

0

$$s_n=\frac{1}{1+x^2} + \frac{1}{(1+x^2)^2} + \dots + \frac{1}{(1+x^2)^n}$$ multiply $s_n$ by $\frac{1}{1+x^2}$ then find $s_n-\frac{1}{1+x^2} s_n$ so $$s_n-\frac{1}{1+x^2} s_n=\\(\frac{1}{1+x^2} + \frac{1}{(1+x^2)^2} + \dots + \frac{1}{(1+x^2)^n})- (\frac{1}{(1+x^2)^2} + \frac{1}{(1+x^2)^3} + \dots + \frac{1}{(1+x^2)^{n+1}})=\\\frac{1}{(1+x^2)} ... 2 Hint: let u=\frac{1}{1+x^2}\in(0,1]. If x=0, then u=1 and your sum is just n. If x\neq 0, then u\in(0,1) and your sum is$$ \sum_{i=1}^nu^i=u\sum_{i=0}^{n-1}u^i=u\frac{1-u^n}{1-u}\to\frac{u}{1-u}\text{ as }n\to\infty. $$2 Just like you would any other geometric series...$$r+r^2+\cdots+r^n=\frac{r^{n+1}-r}{r-1}\qquad \text{ for }r\neq 1$$Plug in r=\left(\dfrac{1}{1+x^2}\right). 0 Without loss of generality, we assume that a\leq b. Since a=2\left(\frac{ab}{2b}\right)\leq2\left(\frac{ab}{a+b}\right)< 2\left(\frac{ab+1}{a+b}\right)<3, we have a=1 or a=2. If a=1, then b can be any natural number and \frac{a^3b^3+1}{a^3+b^3}=1. If a=2, then \frac{2b+1}{b+2}=\frac{ab+1}{a+b}<\frac{3}{2} gives 4b+2<3b+6, ... 0 Try manipulating the first inequality to define a in terms of b (or vice versa)$$ \frac{ab+1}{a+b} < \frac{3}{2} \Rightarrow 2ab+2 < 3a+3b \Rightarrow 2ab - 3a < 3b-2  \Rightarrow a < \frac{3b-2}{2b-3}$$Notice that if a=1, then the second fraction involving a and b would become$$ \frac{(1)^3b^3+1}{(1)^3+b^3} = ...

4

The finite geometric series is given by $$G(q,n)=\sum_{k=1}^nq^k=\frac{1-q^{n+1}}{1-q}$$ Now $\frac{d}{dq}G(q,n)$ is the series you are looking for...

1

Start with the standard formula for the sum of the first $n$ terms of a geometric series and differentiate both sides. This will give you the formula you have

1

We have $$0>\sum_{n\geq1}\left(\frac{1}{\left(z+n\right)^{1/2}}-\frac{1}{n^{1/2}}\right)=-\frac{1}{2}\sum_{n\geq1}\int_{n}^{n+z}\frac{1}{x^{3/2}}dx\geq-\frac{1}{2}\sum_{n\geq1}\int_{n}^{n+1}\frac{1}{x^{3/2}}dx>$$ $$-\frac{1}{2}\sum_{n\geq1}\frac{1}{n^{3/2}}>-\infty.$$ If you prefer, we can use directly the M- test with the same passages ...

0

This can be done with Maple by sol := RealDomain:-solve({-x*y*z+z^3 = 20, -x*y*z+y^3 = 6, x^3-x*y*z = 2}, explicit); $$\left\{ x=-\frac 1 7\,{7}^{2/3},y=\frac 3 7\,{7}^{2/3},z=\frac 5 7\,{7}^{2/3} \right\},\,\left\{ x=-\sqrt [3]{2},y=\sqrt [3]{2},z=2\,\sqrt [3]{2} \right\}$$ L := seq(rhs(sol[j][1])^3+rhs(sol[j][2])^3+rhs(sol[j][3])^3, j = 1 .. 2); ...

2

Hint. You have \left\{ \begin{aligned} x^3=2+t\\ y^3=6+t\\ z^3=20+t \end{aligned} \right. Hence $$x^3 y^3 z^3 = t^3 = (2+t)(6+t)(20+t)$$

2

Your idea of putting $t=xyz$ is quite correct. You then have $x^3=t+6,y^3=t+6,z^3=t+20$, whence $t^3=(t+2)(t+6)(t+20)$. Expanding, you see that $7t^2+43t+60=0$. The rest is easy and I leave it to you. (by the way, the answer is $m+n=158$).

0

Hint: showing normal convergence is usually easier, and a sufficient condition for uniform convergence. (when and if the test for normal convergence fails, then it is time to start thinking about the bad possible situation — that there may be uniform convergence nevertheless.)

1

You have $$\frac{1}{(z+n)^{1/2}}-\frac{1}{n^{1/2}}=\frac{n^{1/2} - (z+n)^{1/2}}{n^{1/2} (z+n)^{1/2}}$$ By the mean value theorem, $(z+n)^{1/2}=n^{1/2} + \frac{\xi_n}{2 n^{1/2}}$, where $\xi_n \in (0,1)$. Also $(z+n)^{1/2} \geq n^{1/2}$. Putting things together, the numerator is at most $\frac{1}{2 n^{1/2}}$ in magnitude while the denominator is at least ...

0

Another way can be this. Let $A_{k}=\sum_{s\leq k}c_{s}$. We have by partial summation $$\sum_{k\leq m}\frac{c_{k}}{nm+k}=\sum_{k\leq m-1}\frac{A_{k}}{\left(nm+k\right)\left(nm+k+1\right)}=\frac{1}{n^{2}}\sum_{k\leq m-1}\frac{A_{k}}{\left(m+k/n\right)\left(m+\left(k+1\right)/n\right)}$$ then $$\left|\sum_{k\leq ... 0 HINT: If n =\prod_{k} p_k^{\alpha_k}, then sum of all divisor of n is$$\prod_{k} \frac{p_k^{\alpha_k+1}-1}{p_k-1}$$. Now, note that 2015 = 5. 13. 31, then you get$$S_n = \frac{5^{n+1}-1}{5-1}\frac{13^{n+1}-1}{13-1}\frac{31^{n+1}-1}{31-1} = \frac{1}{1440} (5^{n+1}-1)(13^{n+1}-1)(31^{n+1}-1)$$. 0 You had the right intuition that S_n would be some sort of geometric series. Note that the sum of the positive factors of 2015^n is given by$$S_n = (1+5+5^2+\cdots+5^n)(1+13+13^2+\cdots+13^n)(1+31+31^2+\cdots+31^n).$$Do you see why? (If you don't, try FOILing that product and see what terms are being added). From this, it follows that ... 3 Let the function f(x) be given by$$f(x)=\left(1-\beta\frac{\log \log x}{\log x}\right)^{\beta}$$where x>1. We wish to find the Taylor series for f(x). To that end, let g(t) be the function given by$$g(t)=(1-t)^{\beta}$$and note that f(x)=g\left(\beta\frac{\log \log x}{\log x}\right). Straightforward calculation shows that the ... 0 A generalization of the \frac{1}{2} proof, where I can get different values like \frac{1}{3} without resorting to anything that is even remotely controversial:$$f[n] = 1 - f[n+1]\underbrace{f[n]+f[n+1]}_{\text{what if not same?}}=1f[n] = 1-1+f[n+2]f[n] = 1-1+1-f[n+3] = 1-f[n+3] = 1 - f[n+1]f[n] = f[n+2]f[n+1] = f[n+3]$$... 0 Well my solution looks basically the same as that of A.G; perhaps it's a little more elementary. I'll include it since I just spent some time on it! Suppose$$\sum_{k=1}^{\infty}|a_k| < \infty, \sum_{k=1}^{\infty}|b_k|^2 < \infty.$$Let f(z) = \sum_{k=1}^{\infty}a_kz^k, g(z) = \sum_{k=1}^{\infty}b_kz^k. Then in the open unit disc,$$f(z)g(z) = ...

1

We need to prove that $$\sum_{k=1}^\infty\left|\sum_{j=1}^ky_j\left(\frac{1}{\lambda}\right)^{k-j}\right|^2\frac{1}{\lambda^2}<+\infty.$$ Denote $x_k=\sum_{j=1}^ky_j\left(\frac{1}{\lambda}\right)^{k-j}$, which gives $$x_{k+1}=\frac{1}{\lambda}x_k+y_{k+1}, \qquad x_0=0,\ k\ge 0.\tag1$$ Do $z$-transform of the equation, i.e. multiply by $z^{k+1}$ and ...

2

The partial sums $C_N$ of the sequence $c_{n\text{ mod } r}$ satisfy $$|C_N| = \left|\sum_{n=0}^Nc_{n\text{ mod }{r}}\right| \leq \sum_{n=0}^{r-1}|c_{n}|$$ and is therefore bounded. Since $a_{n}$ is monotonicly decreasing with $\lim\limits_{n\to\infty} a_n = 0$ we have that Dirichlet's test apply and it follows that the series $\sum c_{n\text{ mod }r}a_n$ ...

-1

Hint $$\lim_{x \rightarrow \infty} \left( 1 - \beta \frac{\log(\log(x))}{\log(x)} \right)^\beta = \lim_{y \rightarrow \infty} \left( 1 - \beta \frac{\log(y)}{y} \right)^\beta = \lim_{z \rightarrow \infty} \Big( 1 - \beta z \exp(-z) \Big)^\beta$$

1

If the real part of $\tau$ is an integer, then $q^2$ is a positive real. If the real part of $\tau$ is a half-integer, then $q^2$ is a negative real. In both cases, $\operatorname{E}_2(\tau)$ must be real. Due to $\operatorname{E}_2$ being holomorphic, infinitesimal deviations from those vertical rays $2\Re\tau\in\mathbb{Z}$ result in non-real changes to ...

3

While your working is admirable, your final step of 'taking limits' is completely undefined, i.e., ask yourself, what are you really doing when you 'take limits' in this context? What definition or process are you referring to? If you are using the usual, elementary calculus notion of 'limit', then you have a divergent limit 'equalling' a divergent limit - ...

2

Note that $$111 = 3 \times 37.$$ Note also that $$a = p, b = p q, c = p q^2.$$ Whence $$3 \times 37 = p \big( 1 + q + q^2 \big).$$ Case $p = 111$ We obtain $$3 \times 37 = 111 \big( 1 + q + q^2 \big) \Rightarrow 1 + q + q^2 = 1 \rightarrow q = 0.$$ Therefore $$(a,b,c) = (111,0,0).$$ But this is no geometric progression. Case ...

0

Hint: Let the numbers in G.P. be $$kr, k, \frac{k}{r}$$ having a common ratio $r$ As per given condition you have $$kr+ k+\frac{k}{r}=111$$

2

You’re taking the product of all products of the form $a_ia_j$ with $0\le j\le i\le n$. For $0\le i\le n$, $a_i$ appears as a first factor $i+1$ times, once for each $j$ satisfying $0\le j\le i$. Each $a_j$ appears as a second factor $n-j+1$ times, once for each $i$ satisfying $j\le i\le n$. Thus, each $a_k$ appears a total of $$(k+1)+(n-k+1)=n+2$$ times, ...

7

HINT: Consider the sequence $\langle x_n:n\in\Bbb N\rangle$ defined by $$x_n=\begin{cases} 0,&\text{if }n\text{ is even}\\ -\frac1n,&\text{if }n\text{ is odd}\;. \end{cases}$$

1

You might be surprised to hear you don't actually have to prove the relation converges before finding what it converges to. Assume $\lim_{n \to {\infty}} x_n=a$ and $\lim_{n \to {\infty}} y_n=b$, also assume that the relation converges, this means $x_{n-1}=x_n$. Substitute... $(1) \quad a=30-\dfrac{b}{2}$ $(2) \quad b=30-\dfrac{a}{2}$ Substitute the ...

1

The author of that proof skips a lot of steps! It seems that the author is secretly making use of the Borel-Cantelli lemma. Equation (*) shows that $\sum_{n=1}^{\infty} Pr[|b_n|\leq |X_n|] < \infty$, which means (by Borel-Cantelli) that, with prob 1, the event $|b_n|\leq |X_n|$ only happens for a finite number of integers $n$. This also means that ...

6

Hint $$y_{n+1}=30-\frac{30-\frac{y_n}{2}}{2}=\frac{60+y_n}{4}$$ Therefore $$y_{n+1}-20=\frac{y_n-20}{4}$$ From here you should be able to show that $$y_n-20=\frac{y_0-20}{4^n}$$

1

$y=\phantom{}_2 F_1(a,b;c;z)$ is the regular solution of the ODE: $$z(1-z)y''+\left[c-(a+b+1)z\right]y'-ab y=0$$ hence $y=\phantom{}_2 F_1(-n,n;1/2;z)$ is the regular solution of the ODE: $$z(1-z)y''+\left(\frac{1}{2}-z\right)y'+ n^2 y=0$$ and $y=\phantom{}_2 F_1\left(-n,n;\frac{1}{2};\frac{1-z}{2}\right)$ is the regular solution of the ODE: ...

0

Another way can be this. We have $$\sum_{n\geq2}\frac{\sqrt{n+2}-\sqrt{n-2}}{n^{p}}=2^{1-p}+\frac{1}{2}\sum_{n\geq3}\frac{1}{n^{p}}\int_{n-2}^{n+2}\frac{1}{\sqrt{t}}dt=2^{1-p}+2\sum_{n\geq3}\frac{1}{n^{p}\sqrt{c_{n}}}$$ where $c_{n}\in\left[n-2,n+2\right]$ by the mean value theorem. Then follows that $$... 2 I post this answer because Dirichlet's test has not been mentioned in any of the previous answers. Let a_n=(-1)^{n(n-1)/2} and b_n=1/n. The partial sums of a_n are bounded and b_n is decreasing and converging to 0. Dirichlet's test implies the series is convergent. 1 Let's start off with the case of the Fibonacci sequence. We have$$\sum_{n = 1}^k \left( \frac{F_{n+2}}{F_{n+1}} -\frac{F_{n+3}}{F_{n+2}} \right) = \left(\frac{F_3}{F_2} - \frac{F_4}{F_3}\right) + \left(\frac{F_4}{F_3} - \frac{F_5}{F_4}\right) + \cdots + \left(\frac{F_{k+2}}{F_{k+1}}- \frac{F_{k+3}}{F_{k+2}}\right)\\ = \frac{F_3}{F_2} - ...

1

Gauss's Test asserts that for a positive series with terms $a_n$, if the ratio of successive terms $$\frac{a_n}{a_{n+1}}=1+\frac{h}{n}+\frac{B(n)}{n^{\alpha}}$$ where $\alpha>1$ and $B(n)$ is bounded as $n\to \infty$, then the series converges if $h>1$ and diverges otherwise. For the problem of interest, ...

2

Note $$\sum_{m = 1}^n \frac{1}{m} = \log n + \gamma + \varepsilon_n,$$ where $\gamma$ is Euler constant, $\lim \varepsilon_n = 0$. Hence the general term is $$n^r \exp(-k(\log n + \gamma + \varepsilon_n)) = \exp(-k(\gamma + \varepsilon_n))\frac{1}{n^{k - r}} \sim \exp(-k\gamma)\frac{1}{n^{k - r}}.$$ Therefore the series converges when $k - r > 1$.

0

@RRL provided a solid approach so I thought it might be instructive to present a different way forward. Here we use asymptotic analysis. To that end, we write $$\sqrt{n\pm2}=n^{1/2}\left(1\pm\frac2n\right)^{1/2}=n^{1/2}\left(1\pm\frac1n+O\left(\frac{1}{n^2}\right)\right)$$ Thus, ...

0

Let $A_N$ be the $N$th partial sum of $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}.$ Let $B_N$ be the $N$th partial sum of $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{2n}.$ Let $C_N$ be the $N$th partial sum of $\sum_{n=1}^{\infty}\frac{(-1)^{n(n-1)/2}}{n}.$ Then $C_{2N} = A_N + B_N.$ By the alternating series test, both $A_N,B_N$ converge. Hence $C_{2N}$ ...

4

Some of the posted answers are using rearrangements or regrouping, but these are treacherous techniques to apply to series which diverge absolutely (as this one plainly does). Leibniz' Theorem as usually stated only applies to alternating series, but we can modify the proof of Leibniz' Theorem to cover this particular series. Leibniz Theorem: A series of ...

2

Hint: $$\frac{\sqrt{n+2}-\sqrt{n-2}}{n^p} = \frac{4}{n^p(\sqrt{n+2}+\sqrt{n-2})} \sim 2n^{-(p + 1/2)}$$

3

After $\ n = 1 \$ , the exponents of (-1) are binomial coefficients which are "double-alternating" between even and odd integers. So the series looks like $$1 \ - \frac{1}{2} \ - \ \frac{1}{3} \ + \ \frac{1}{4} \ + \ \frac{1}{5} \ - \ \frac{1}{6} \ - \ \frac{1}{7} \ + \ \ldots \ \ .$$ As lulu I think properly objected to my separation of terms ...

-1

after the theorem of Leibniz is this series convergent since $$\lim_{n \to \infty}\frac{1}{n}=0$$ and $$\frac{1}{n+1}\le \frac{1}{n}$$ for all $n$

6

The parity of $\frac{n(n-1)}{2}$ is 4-periodic. Thus the sequence $(-1)^{\frac{n(n-1)}{2}}$ equals to: $$1, \, -1, \, -1, \, 1, \, 1, \, -1, \, -1, \, 1, \, 1, \, -1, \, -1, \, 1 , \cdots$$ The original series' partial sum truncated at $N$ equals to $$\sum_{k=0}^{K} \left( \frac{1}{4k+1} - \frac{1}{4k+2} - \frac{1}{4k+3} + \frac{1}{4k+4}\right) + ... 0 Easiest is to define:$$ A(z) = \sum_{n \ge 0} a_n z^n $$then shift indices so there aren't subtractions there:$$ a_{n + 2} = 4 a_{n + 1} - 4a_n + \binom{n + 2}{2} 2^{n + 2} + 1 $$Multiply by z^n and sum over n \ge 0:$$ \sum_{n \ge 0} a_{n + 2} z^n = 4 \sum_{n \ge 0} a_{n + 1} z^n - 4 \sum_{n \ge 0} a_n z^n + \sum_{n \ge 0} \binom{n + ...

3

It's uncountable. Take any real number between 0 and 1 and make a sequence of your type out of the decimal as in: $$\alpha = .912092013.... \;\;\;a_1 = 9 \;a_2=91\;\;a_3=912\;\;a_4 =9120$$ And so on. Clearly increasing and there is only one $a_i$ between any two consecutive powers of 10. (in case of ambiguity for terminating rationals, use the expansion ...

1

You want to use a comparison test. Remember with series that you can disregard a finite number of terms. When $n \geq 2$ $$n^7+13n^5+9n+2 \leq n^7 +13n^7 + 9n^7+n^2 = 24n^7,$$ which implies $$\frac{1}{n^7+13n^5+9n+2} \geq \frac{1}{n^7 +13n^7 + 9n^7+n^2} \geq \frac{1}{24n^7}.$$ For the numerator, you have $$n^6 + 13n^5 + n + 1 \geq n^6.$$ Therefore ...

5

Use the limit comparison test. Let $\displaystyle a_n=\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}$ and $\displaystyle b_n=\frac{n^6}{n^7}=\frac{1}{n}$. Since \begin{align}\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}&=\lim_{n\to\infty}\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}\cdot\frac{n}{1}\\ ...

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