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Take any sequence $s_n$ such that $s_n \to 1$ as slowly as you want, with $s_0 = 0$. Let $a_n = s_n - s_{n-1}$. Then the partial sums $\sum_{i=1}^n a_i = s_n$.

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Hint: $\sin (1/n) \leqslant 1/n$ Note: this hint was for the OP's original question, which was about the series $$\sum_{n=2}^\infty {\frac{e^{\frac{1}{n}}\sin(\frac{1}{n})}{{\sqrt{n}}}}$$

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$$\frac{1}{\sqrt{2k}+\sqrt{2k+2}}=\frac{\sqrt{2k+2}-\sqrt{2k}}{2}$$ The sum telescopes to $\displaystyle \frac{1}{\sqrt n} \frac {\sqrt{2n+2} - \sqrt 2 }{2}$ which converges to $1/\sqrt{2}$ If you miss this, you can still use integrals to get bounds on $\sum_{k=1}^n \frac{1}{\sqrt{2k}+\sqrt{2k+2}}$

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If one attempts to use Cauchy's First Limit Theorem, then we start with \begin{align} \lim_{n\to \infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}}&=\lim_{n\to \infty}\sqrt{n}\frac1n\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}} \tag 1\\\\ \end{align} The Theorem reveals that since $\lim_{n\to ... 4 Hint. One may just interchange sum and integration : $$\int_0^\infty\sum_{n=0}^{\infty}\frac{x^n}{2^{(n+1)^sx^{n+1}}+1}dx=\sum_{n=0}^{\infty}\int_0^\infty\frac{x^n}{2^{(n+1)^sx^{n+1}}+1}dx$$ then, by the change of variable $$u=(n+1)^sx^{n+1}, \quad du=(n+1)^{s+1}x^ndx,$$ one gets $$... 4 The proof is simple: partition the sum to whenever \pi(i) jumps up by 1, i.e. when i=P_j , the j'th prime. Then your sum is just:$$2+\sum_{j=1}^\infty\left[\frac{2n}{j+n+1}\right](P_{j+1}-P_j)=2+\sum_{j=1}^{n-1}(P_{j+1}-P_j)=P_n.$$Where we noticed that the floored term is always 1 until j exceeds n-1, where it becomes 0. So your sum telescopes ... 3 If we want to solve this without any guessing and without using induction in any way, the best way to do so is to work with generating functions. This method, given that it really makes no assumptions on the final form, takes considerably more work, but can also solve generalizations of this problem. In particular, the big trick is to consider the ... 3 This is a classic partial sums of integrals problem. Your expression is:$$\frac{1}{\sqrt{n}}\sum\limits_{k=1}^n \frac{1}{\sqrt{2k}+\sqrt{2k+2}}$$write as$$\frac{1}{n}\sum\limits_{k=1}^n \frac{1}{\sqrt{2\frac{k}{n}}+\sqrt{2\frac{k}{n}+2}}$$then the limit is the same as the integral$$\int_0^1\frac{1}{\sqrt{2x}+\sqrt{2x+2}}dx= ... 2 This is similar to the difference between integral and Cauchy principal value For example you know that $$\int _ {\mathbb R} x dx$$ does not exists, but $$\lim_{R \to \infty} \int_{-R}^R x dx = 0$$ which is the Cauchy principal value. The main point is that$\int_\mathbb R$is not defined as a limit$\lim_{R \to \infty} \int_{-R}^R$.. For the riemann ... 2 Hint: $$\frac{\displaystyle \frac{(2n+2)!}{4^{n+1}((n+1)!)^2(2n+3)}x^{2n+3}} {\displaystyle\frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1}} = \frac{(2n+1)(2n+2)(2n+1)}{4(n+1)^2(2n+3)}x^2 = \frac{(2n+1)^2}{(2n+2)(2n+3)}x^2\to x^2.$$ 2 You can do$\frac 6{\pi^2} \sum_{n=1}^\infty \frac 1{n^2}$or$\sum_{n=1}^\infty \frac 1{n(n+1)}$2 Well,$\left\lfloor \frac{\phi(n)}{n-1}\right\rfloor$is just an extraordinarily fancy way to say "1 when$n$is prime and 0 otherwise". The rest is kinda trivial. Indeed, non-zero terms of this sum would correspond to the primes, and the values of$\pi(n)$for them are just consecutive positive integers. $$\sum_{n=2}^{\infty}\frac{1}{\pi^s(n)}\left\lfloor ... 1 With no induction then I guess what you want is solve it "directly". To me the first choice would be with ansatz, and the correct one to use is a_n = \alpha n + \beta. The reason can be argued by dominant balance. If a_n \sim n, then S_n \sim n^2/2, so LHS and RHS are balanced (both 2n^2). If we assume If a_n \sim n^2, then S_n \sim n^3/3, this ... 1 Rearranging the given equation$$4S_n=n(a_n+a_{n+1})$$gives$$\frac {S_n}n=\frac 12 \left(\frac{a_n+a_{n+1}}2\right)$$i.e. the average of the first n terms is half the average of the n-th and (n+1)-th terms, which is almost linear. This leads us to suspect that the sequence might be an arithmetic progresion (AP). We have been given that the first ... 1 Hint: Check that \lim_{n \to +\infty} \frac{a_1+\cdots + a_n}{n} = \lim_{n \to +\infty}a_n. With this, apply the Leibniz alternating test. 1 The polylogarithm function is defined as$$\textrm{Li}_{s}\left(z\right)=\sum_{k\geq1}\frac{z^{k}}{k^{s}} $$for all complex s and for \left|z\right|<1 . So observe that$$\frac{\partial}{\partial s}\left(\textrm{Li}_{s}\left(\frac{1}{e}\right)\right)=-\sum_{k\geq1}\frac{\log\left(k\right)}{e^{k}k^{s}} $$and so$$\frac{\partial}{\partial ... 1$U_{n+1}-\frac{5}{2}=\frac{1}{5}(U_n-\frac{5}{2})$Classic trick for arithmetico-geometric series. (Solve$U_{n+1}-\alpha = \frac{1}{5}(U_n-\alpha)$) 1 You made a mistake. There shouldn't be any negative sign. If you do the calculation correctly, you will get $$x_{n+1} = \phi(x_n) = \frac{10}{3}x_n$$ so that$\phi'(x)=\frac{10}{3}>1, \forall x$. Edit: remember, when you study the convergence of a fixed point method, it is not the iteration map$\phi$that you want to be bounded by 1 (in modulus), ... 1 The simple induction would work with no problem if we were given that$\sum a_k\le 1/2$. Which is enough for the presumed application to infinite products, but doesn't solve the current problem. Hint for the problem as given: $$\log(1+t)\le t\quad(t\ge0)$$ and $$\log(1+2t)\ge t\quad(0\le t\le 1).$$ (You can prove both these by writing the logarithm as the ... 1 Note that${(2n)!\over4^n(n!)^2}$can be expanded and then grouped in two different ways: $${(2n)!\over4^n(n!)^2}=\left((2n)(2n-1)\over(2n)(2n) \right)\left((2n-2)(2n-3)\over(2n-2)(2n-2) \right)\cdots\left(4\cdot3\over4\cdot4 \right)\left(2\cdot1\over2\cdot2 \right)$$ and$\${(2n)!\over4^n(n!)^2}={1\over2n}\left((2n)(2n-1)\over(2n)(2n-2) ...

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