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8

Try bounding $$E_n = \left ( 1 + \frac{1}{n} \right )^n - \sum_{k=0}^n \frac{1}{k!}$$ using the binomial expansion you wrote, and pairing off terms. That is, write $$E_n = (1-1) + (1-1) + (1/2 - 1/2n - 1/2) + \dots$$ Using the factorial definition of the binomial coefficient will prove useful.


7

If a sequence is convergent, then its limit is unique, and every subsequence converges to that limit. Can you find two subsequences of your sequence that have a different limit?


6

There is no need to estimate/bound the difference between $\displaystyle\;\left(1 + \frac{1}{n}\right)^n\;$ and $\displaystyle\;\sum_{k=0}^\infty \frac{1}{k!}\;$. In the expansion $$\left(1+\frac{1}{n}\right)^n = \sum_{k=0}^n \frac{n!}{k!(n-k)!}\frac{1}{n^k} =\sum_{k=0}^\infty \frac{\alpha_{n,k}}{k!}\quad\text{with}\quad\alpha_{n,k} = \begin{cases} ...


5

Simply a heads-and-tails game in disguise... Consider a heads-and-tails game such that the probability of success of the $n$th throw is $$p_n=\frac{na}{1+na}.$$ Then the probability that the $n$th throw is the first success is $$ p_n\prod_{k=1}^{n-1}(1-p_k)=na\prod_{k=1}^n\frac1{1+ka}, $$ hence $$ ...


4

Let $a_n = \left(1+ 1/n\right)^n$. By the binomial theorem, $$a_n = 1 + 1 + \frac1{2!}\left(1- \frac1{n}\right)+ \ldots +\frac1{n!}\left(1- \frac1{n}\right)\ldots\left(1- \frac{n-1}{n}\right)\leq \sum_{k=0}^{\infty}\frac1{k!}=e,$$ The sequence $a_n$ is increasing and bounded, so it converges. Hence, $$\lim_{n\rightarrow \infty} a_n \leq e.$$ For $m < ...


2

The series seems to follow the pattern $a, a+1, 2a$ if we remove $8$.


2

Your proof looks fine. This is how I would prove it. If $a>1$, put $x_n=\sqrt[n]{a}-1$. Then $x_n>0$, and, by the binomial theorem, $$1+nx_n\le \left(1+x_n\right)^n=a$$ so that $$0<x_n\le \frac{a-1}{n}$$ Hence $x_n\rightarrow 0$. If $a=1$ this is trivial, and if $0<a<1$, the result is obtained by taking recipocals!


2

The function $f(x) = 1+\frac1{1+x}$ is decreasing on $\mathbb{R}^+$; $0\lt a\lt b\implies f(b)\lt f(a)$. From this you can 'hopscotch' back and forth between odd and even terms: since $a_1\lt a_3$, then $a_2=f(a_1)\gt a_4=f(a_3)$, and therefore $a_3=f(a_2)\lt a_5=f(a_4)$, etc. Now, consider the two subsequences $b_n = a_{2n}$ and $c_n=a_{2n+1}$. You've ...


2

I do not think that they are simplified forms in the reverse of a polylogarithm but you always can to build them by identification of series and series compositions. Just write $$\dfrac{1}{\text{Li}_s(z)}=\sum_{i=-1}^{\infty}a_i z^i$$ (the summation must start at $i=-1$ because the expansion of the polylogarithm starts with a first term equal to $z$) and so ...


2

You can conclude if the series converges or not,using the ratio test. It is like that: $a_n=\frac{1}{n^n}$ $$\frac{a_{n+1}}{a_n}=\frac{\frac{1}{(n+1)^{n+1}}}{\frac{1}{n^n}}=\frac{n^n}{(n+1)^{n+1}} \to 0<1$$ So,the series $\sum_{n=1}^{\infty} \frac{1}{n^n}$ converges.


1

Use that $\sum\limits_{k=1}^{\infty} a_k $ converges if and only if $\sum\limits_{k=j}^{\infty} a_k $ converges for a $j \in \mathbb N$ Edit: Here is the full solution: $\sum\limits_{n=1}^{\infty} \frac{1}{n^n}<\infty \iff \sum\limits_{n=2}^{\infty} \frac{1}{n^n}<\infty$ and $ \sum\limits_{n=2}^{\infty} ...


1

We have $$e-\left(1+\frac1n\right)^n=e-\exp\left(n\ln\left(1+\frac1n\right)\right)\sim_\infty e-\exp\left(n\left(\frac1n-\frac1{2n}\right)\right)=e-e\exp\left(-\frac1{2n^2}\right)\\\sim_\infty e-e\left(1-\frac{1}{2n}\right)=\frac e{2n}$$ so the given series is divergent by comparison with the harmonic series $\frac e2\sum\limits_n\frac1n$.


1

Hint Do a Limit Comparison with $\sum\limits_{n=1}^\infty \frac{1}{2^n}$. Remark: The intuition is that for $n$ large, the number $\frac{5n^3+1}{n^3+n+1}$ is close to $5$. So for large $n$, our terms behave like $\frac{5}{2^n}$, which goes down plenty fast enough for convergence. For details, divide our $n$-th term by $\frac{1}{2^n}$. After some ...


1


1

$$\dfrac{5n^3+1}{2^n\left(n^3+n+1\right)}\leqslant\dfrac{5}{2^n}.$$


1

You can use Pell number in your proof (two steps): 1)Show that $\displaystyle a_n=\frac{P_{n}+P_{n-1}}{P_{n}}$. 2)Solve reccurence $P_{0}=0$, $P_{1}=1$, $P_{n}=2P_{n-1}+P_{n-2}$ to find direct formula for $P_{n}$. Both steps you can do for example by induction.


1

Clearly $a_n>1$. Note $$|a_{n+1}-\sqrt{2}|=\frac{\sqrt{2}-1}{1+a_n}|a_n-\sqrt{2}|\le\frac{\sqrt{2}-1}{2}|a_n-\sqrt{2}|.$$ Thus $$|a_n-\sqrt{2}|\le \left(\frac{\sqrt{2}-1}{2}\right)^{n-1}|a_1-\sqrt{2}|$$ and hence $$\lim_{n\to\infty}a_n=\sqrt{2}.$$


1

Note that on $[1,2]$ the function $f(x) = 1 + {1 \over 1 + x}$ satisfies $|f'(x)| \leq {1 \over 4}$. So by the mean-value theorem, for $x$ and $y$ in the interval you have $|f(x) - f(y)| \leq {1 \over 4}|x - y|$. Letting $x = a_n$ and $y = a_{n+1}$ this implies that $$|a_{n+2} - a_{n+1}| \leq {1 \over 4}|a_{n+1} - a_n|$$ From here it's pretty fast that the ...


1

Most inelegant method I guess: $a_{n+2}=1+\cfrac{1}{1+a_{n+1}}=1+\cfrac{1}{1+1+\cfrac{1}{1+a_n}}=1+\cfrac{1+a_n}{3+2a_n} $ So $a_{n+2}-a_n=2\cfrac{4-a_n^2}{3+2a_n}$ Clearly from definition $a_n>1$ Also if $a_n>\sqrt{2}$ then $a_{n+2}>1+\cfrac{1+\sqrt{2}}{3+2}=\cfrac{6}{5}+\cfrac{\sqrt{2}}{5}>\cfrac{4\sqrt{2}}{5}+\cfrac{\sqrt{2}}{5}=\sqrt{2}$ ...


1

Let $k$ be an integer, $|\sin((2k)\pi/2) - \sin((2k + 1)\pi/2)| = 1$. Thus when $\epsilon = 1$, for all $N$ we can find elements in the sequence which differ by $\epsilon$. Let $k = N$. This contradicts the definition of convergence.


1

Well, $x^2 - 2 y^2 = \pm 1.$ We want to know about primes dividing $x,$ which will then divide $2 y^2 \pm 1.$ If prime $p \equiv 5 \pmod 8,$ then $2y^2 - 1 \equiv 0 \pmod p$ is impossible, as $2y^2 \equiv 1 \pmod p$ implies $(2|p) = 1,$ which is false. Also $2y^2 + 1 \equiv 0 \pmod p$ is impossible, as $2y^2 \equiv -1 \pmod p$ implies $(-2|p) = 1,$ which ...


1

Apart from fact that we need to specify that $n\ge 2$, the argument is correct for $a\gt 1$. It would be simpler to use the fact that if $n\ge 1$ then $a\ge 1+nh$. However, the conclusion that we can find a suitable $m$ for all $a\gt 0$ has not been proved. For the argument, though correct when $h$ is positive, does not work if $h\lt 0$. For the case $0\lt ...


1

You could just as well remove the 14 or remove the 30.


1

Hint: $$\cos\frac{r\pi}{6} = \text{Real}\left(e^{\dfrac{i\pi}{6}}\right)^r$$


1

$\displaystyle\cos\frac{r\pi}6=$Real $(e^{\dfrac{ir\pi}6})$ So, $\displaystyle\sum_{r=0}^{12}\binom{12}r\cos\frac{r\pi}6$=Real $\displaystyle\left(\sum_{r=0}^{12}\binom{12}re^{\dfrac{ir\pi}6}\right)$ Now $\displaystyle\sum_{r=0}^{12}\binom{12}re^{\dfrac{ir\pi}6}=\sum_{r=0}^{12}\binom{12}r(e^{\dfrac{i\pi}6})^r=(1+e^{\dfrac{i\pi}6})^{12}$ and ...



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