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11

Clearly, $$ \frac{k}{n^2+k^2}\ge \int_k^{k+1}\frac{x\,dx}{n^2+x^2}, $$ and hence $$ \sum_{k=1}^\infty\frac{k}{n^2+k^2}\ge \int_1^\infty\frac{x\,dx}{n^2+x^2} = \left.\frac{1}{2}\log(x^2+n^2)\right|_1^\infty=\infty. $$ On the other hand \begin{align} ...


5

If the sum is $(-3)^{n-1}$ then you cannot factor out a -1. Since $(-3)^{n-1} = (-1)^{n-1}(3)^{n-1}$ There would be no such thing as an alternating series if you could just move the negative sign out in front of the sum!


5

Use the laws of exponents: $$\frac{7^{2n}}{4^{3n}} = \frac{\left(7^2\right)^n}{\left(4^3\right)^n}= \frac{49^n}{64^n}=\left(\frac{49}{64}\right)^{n}$$


4

$$\begin{align} \sum_{n=1}^{\infty } \frac{-3^{n-1}}{7^{n}} & = - \frac{1}{7} \sum_{n=1}^{\infty } (\frac{3}{7})^{n-1} \\ & = - \frac{1}{7} \frac{1}{1-\frac 3 7} \\ & = - \frac 1 4 \\[2ex] \sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}} & = \frac{1}{7} \sum_{n=1}^{\infty } (-\frac{3}{7})^{n-1} \\ & = \frac{1}{7} \frac{1}{1+\frac 3 7} \\ ...


3

You could get away with a simple estimate of the expected range of the limit: With $S_1(n)=\sum_1^{n}\frac{k}{n^2+n^2}$ and $S_2(n)=\sum_{1}^{n}\frac{k}{n^2}$, we have $$S_1(n)<S(n)<S_2(n)$$ The limits for $S_1$ and $S_2$ are ...


2

The general term is equivalent to $\frac{1}{k}$, so I don't think this converges at all.


2

For students who feel weak with exponent manipulation: $a$ is the initial term. So just evaluate it at $n=0$ and you have $a$. (It's homework so I'm not giving the result.) $r$ is the constant ratio between terms. So write down the first few terms by evaluating at $n=0$, $n=1$, and just to confirm, at $n=2$. See what ratio you are multiplying by at each ...


2

$$-\frac{1}{3} \sum_{n=1}^{\infty} \frac{3^n}{7^n}=-\frac{1}{3} \sum_{n=1}^{\infty} \left ( \frac{3}{7} \right )^n=-\frac{1}{3} \sum_{n=0}^{\infty} \left ( \frac{3}{7} \right )^n+\frac{1}{3}=-\frac{1}{3} \frac{1}{1-\frac{3}{7}}+\frac{1}{3} \\ =-\frac{1}{3}\frac{7}{7-3}+\frac{1}{3}=-\frac{7}{12}+\frac{4}{12}=\frac{-1}{4}$$ EDIT: $$-\frac{1}{3} ...


2

The expression on the right of (3) can be transformed into \begin{align} \frac{z^n-w^n}{z-w}-nw^{n-1}&=z^{n-1}\frac{1-\left(\frac{w}{z}\right)^{n}}{1-\frac{w}{z}}-nw^{n-1}=\\ &=z^{n-1}\left[1+\frac{w}{z}+\left(\frac{w}{z}\right)^{2}+\ldots+\left(\frac{w}{z}\right)^{n-1}\right]-nw^{n-1}=\\ ...


1

Is induction on $x$ acceptable? Because: $\sum_{i = 0}^{\infty}\left[\sum_{j=0}^{i}\left[\frac{(-1)^j}{j!} \right] \prod_{j=0}^{i}\left[ (x+1-j)\right] \right] =\sum_{i = 0}^{\infty}\left[\sum_{j=0}^{i}\left[\frac{(-1)^j}{j!} \right] \prod_{j=-1}^{i}\left[ (x-j)\right] \right] =\sum_{i = 0}^{\infty}\left[\sum_{j=0}^{i}\left[\frac{(-1)^j}{j!} \right] ...


1

Since $F_n=cF_{n-1}+eF_{n-2}$, if the greatest common divisor of $F_n$ and $F_{n-1}$ does not divide $e$, it must divide $F_{n-2}$. So $f$ divides both $F_{n-1}$ and $F_{n-2}$ but not $e$, hence it must divide $F_{n-3}$. By going backwards, $f$ must divide both $F_1$ and $F_2$, hence it must divide their greatest common divisor. This proves $f|d$.


1

I think that your formula is a particular case of the Newton interpolation formula: Let $f(n)$ a fonction on $\mathbb{N}=\{0,1,\cdots\}$, put $\displaystyle a_n=\sum_{k=0}^n(-1)^k \binom{n}{k}f(n-k)$, and $\displaystyle g(x)=\sum_{n\geq 0}a_n\binom{x}{n}$, then we have $g(m)=f(m)$ for all $m \in \mathbb{N}$. (Also, look at your formula: for $i=0$, we get ...


1

\begin{align*} \sum_{i=0}^\infty \sum_{j=0}^i \frac{(-1)^j}{j!} \prod_{k=0}^i (x-k) &= \sum_{i=0}^{\color{red}{x-1}} \sum_{j=0}^i \frac{(-1)^j}{j!} \prod_{k=0}^i (x-k) \\ &= x! \sum_{i=0}^{x-1} \sum_{j=0}^i \frac{(-1)^j}{j!(x-i-1)!} \\ &= x! \sum_{i=0}^{x-1} \sum_{j=0}^{x-i-1} \frac{(-1)^j}{j!\,i!} &&\text{(reindex $i:=x-i-1$)} \\ ...


1

Hint: Note that $\displaystyle\sum_{j=0}^{\infty}(j-2)a_{j}x^{j}=(x\frac d{dx}-2)\sum_{i=0}^{\infty}a_{i}x^{i}$



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