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6

Just pick a subsequence such that $a_{n_k} > k$, and then let $b_{n_k} = 1/k^2$, and $b_n = 0$ if $n$ is not an index in that subsequence.


5

First of all, since $f$ is continuous, $f(0) = 0$, so $a_0 = 0$. Now, think about $f(x)/x = \sum_{n=0}^\infty a_{n+1}x^n$, which is again continuous. For any $x_n$, we have $f(x_n)/x_n = 0$, so $f(x)/x \rightarrow 0$, as $x \rightarrow 0$. Hence, $a_1 = 0$. Continuing this way, we show that $a_n = 0$ for all $n$. Note that we need to assume that all $x_n$ ...


5

Another way to do is just use basic calculus without using the digamma function: Let $$ f(x)=\sum_{n=1}^\infty\frac{2}{n(3n-1)}x^{3n}. $$ Clearly $\sum_{n=1}^\infty\frac{2}{n(3n-1)}=f(1)$. Note $$ f'(x)=6\sum_{n=1}^\infty\frac{1}{3n-1}x^{3n-1},f''(x)=6\sum_{n=1}^\infty x^{3n-2}=\frac{6x}{1-x^3}. $$ So \begin{eqnarray} ...


4

Using $$ \frac{1}{p(n)} = 2 \left( \frac{1}{n-\tfrac{1}{3}} - \frac{1}{n} \right) $$ and the definition of the digamma function: $$ \sum_{n=1}^\infty \frac{1}{p(n)} = -2 \left( \psi\left(\frac{2}{3}\right) + \gamma \right) = 3 \ln(3) - \frac{\pi}{\sqrt{3}} \approx 1.48204 $$ The value for the $\psi\left(\frac{2}{3}\right)$ can be derived from ...


3

If $f(x)$ is not identically $0$, some $a_n$ is nonzero. Suppose $a_m$ is the first nonzero coefficient. Then $g(x) = f(x)/x^m = a_m + a_{m+1} x + \ldots$ is a convergent power series, therefore continuous, and $g(0) = a_m \ne 0$. By continuity, there is $\epsilon > 0$ such that $g(x) \ne 0$ for $|x| < \epsilon$, and thus $f(x) \ne 0$ for $0 < |x| ...


3

By continuity of $f$ we have that $\,f(0)=\lim_{n\to\infty}f(x_n)=0$. As $f\in C^\infty(-R,R)$, there exist $y_n\in(x_{n+1},x_n)$, such that $$ 0=\frac{f(x_{n+1})-f({x_n})}{x_{n+1}-x_n}=f'(y_n), $$ due to Mean Value Theorem. Clearly $y_n\to 0$, and due to continuiuty of $f'$, we have $f'(0)=\lim_{n\to\infty}f'(y_n)=0$. Similarly, we may found a ...


2

Basically, you don't. The only way to manipulate them is if you know precisely what you're doing and using renormalization techniques or analytic continuation or something, but those techniques won't be taught for awhile and have specific rules about them. Long story short, you don't manipulate them.


2

$x^2$ is positive $$x^2<x^2+4$$ $$0<4$$ Converges for all x


2

$$ \sum_{n,m=0}^\infty A(n,m,n+m) = \sum_{k=0}^\infty\sum_{n=0}^k A(n,n-k,k) $$ $$ \begin{array}{c|ccccccccc} & m=0 & m=1 & m=2 & m=3 & \cdots \\ \hline n=0 & \bullet & \bullet & \bullet & \circ & \cdots \\ n=1 & \bullet & \bullet & \circ & \cdots \\ n=2 & \bullet & \circ & \cdots \\ n=3 ...


2

Since the sum is absolutely convergent, we can sum in any order. In the fourth line, we just sum along the diagonals $n+m = \text{constant}$. This is known as a Cauchy product: http://en.wikipedia.org/wiki/Cauchy_product


2

To show that $e^{2\pi i/n}\ne 1$ if $n\gt 1$, it is enough to observe that $\cos(\phi)=1$ only when $\phi$ is a multiple of $2\pi$. Remark: As has been pointed out in an unfortunately currently deleted answer by lhf, the sum of the roots of a monic polynomial $x^n+ax^{n-1}+\cdots$ is $-a$. So if $n\gt 1$ then the sum of the roots of $x^n-z$ is $0$. No ...


2

This should give the idea to prove (a) and (b). $$\sum_{n=1}^\infty\frac1{n^2}<1+\sum_{n=2}^\infty\frac1{n^2-1}=1+\frac12\sum_{n=2}^\infty\left(\frac1{n-1}-\frac1{n+1}\right)=1+\frac12\cdot\frac32$$ For the latter step, note the telescopic series.


1

Why don't you use the ratio test? The ratio test tells you that the (power) series $\sum a_n$ converges if \begin{equation} \lim_{n \to \infty} \Big| \frac{a_{n+1}}{a_n} \Big| < 1 \end{equation} \begin{equation} \lim_{n \to \infty} \Big| \frac{\frac{e^{n+1}+1}{e^{2n+2}+n}x^{n+1}}{\frac{e^n+1}{e^{2n}+n}x^n} \Big| < 1 \end{equation} \begin{equation} ...


1

Expanded HINT: For one direction, assume (a), and let $A$ be an infinite subset of $V$. Since $A$ is infinite, it contains a sequence $\langle a_n:n\in\Bbb Z^+\rangle$ whose points are all distinct. (That is, $a_m\ne a_n$ if $m\ne n$.) By (a) this sequence has a subsequence $\langle a_{n_k}:k\in\Bbb Z^+\rangle$ that converges to some $x\in V$. Use the ...


1

Hint: Since $Y^k\nu^k=(Y\nu)^k$, our sum, apart from the missing term $k=0$, is the Binomial Theorem expansion of $(Y\nu +1-\nu)^s$. So our sum is $(Y\nu+1-\nu)^s-(1-\nu)^s$.


1

Let $\xi \in \mathbb{R}$. Then $\sum_{n=1}^{\infty}\xi^{n}$ converges iff $|\xi| < 1.$ Since $$0 \leq \frac{x^{2}}{x^{2} + 4} < 1$$ for all $x \in \mathbb{R}$, it follows that $\sum_{n=1}^{\infty}(\dfrac{x^{2}}{x^{2} + 4})^{n}$ converges for all $x \in \mathbb{R}.$


1

We have $a_{n+1} = 1 + \frac{1}{1+a_{n}} = \frac{2+a_n}{1+a_{n}} $. Therefore, $a_n > 1$ for all $n$. Also, $\begin{array}\\ a_{n+1}-\sqrt{2} &= \dfrac{2+a_n}{1+a_{n}}-\sqrt{2}\\ &= \dfrac{2+a_n-\sqrt{2}(1+a_n)}{1+a_{n}}\\ &= \dfrac{2-\sqrt{2}-a_n(\sqrt{2}-1)}{1+a_{n}}\\ &= \dfrac{(\sqrt{2}-1)(\sqrt{2}-a_n)}{1+a_{n}}\\ \end{array} $ so ...


1

Assume contrary that for all $(b_n )\in \ell_1 $ the series $$\sum_{n=1}^{\infty } a_n b_n $$ is convergent. Define $$T_n :\ell_1 \to \mathbb{R} $$ $$T_n (b) =\sum_{j=1}^n a_j b_j .$$ The functionals $T_n $ are linear and continuous and $$\lim_{n\to \infty} T_n b$$ exists for any $b\in \ell_1 $ moreover $||T_n ||=\sup_{1\leq k\leq n} |a_k |$. Thus by ...



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