$$ { ${\frac{1}{n-1+x_1}+...+\frac{1}{n-1+x_n}\le 1}$ } $$
$$ {} $$

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$$u_{n+1}-u_n=\frac{1+2+...+n+(n+1)}{n+3}-\frac{n+1}{2}-\frac{1+2+...+n}{n+2} + \frac{n}{2} $$

Check if the sequence converges?

But how can find this difference? u_{n+1}-u_n=\frac{1+2+...+n+(n+1)}{n+3}-\frac{n+1}{2}-\frac{1+2+...+n}{n+2} + \frac{n...

A question about the digamma function and a recursion

The recursion is T_n=\frac{b}{n+b}+T_{n-1} and T_1=1 . When expanding the recursion we have T_n=\frac{b}{b+n-1}+\frac{b}{b+n-2}+...+\frac{b}{b+1}+1 . However, the solution shows T_n=b({\psi _0}(b + n) - {\psi _0}(b)) where {\psi _0} is the digamma function. The solution indicates {\psi _0}(b + n) - {\psi _0}(b)=\frac{1}{b+n-1}+\frac{1}{b+n-2}+...+\frac{1}{b+1}+\frac{1}{b} . For instance, when n=4,b=1 , {\psi _0}(5) - {\psi _0...
$$S_n = 2(1-\frac{1}{2} + \frac {1}{2} - \frac{1}{3} + .... + \frac {1}{n} - \frac {1}{n+1}) = 2(1 - \frac {1}{n+1}) $$

Sum of reciprocals of the triangle numbers

Consider the sum of n terms : S_n = 1 + \frac{1}{1+2} + \frac {1}{1+2+3} + ... + \frac {1}{1+2+3+...+n} for n \in N . Find the least rational number r such that S_n < r , for all n \in N . My attempt : S_n = 2(1-\frac{1}{2} + \frac {1}{2} - \frac{1}{3} + .... + \frac {1}{n} - \frac {1}{n+1}) = 2(1 - \frac {1}{n+1}) Now what to do with that ' r ' thing ? How to proceed ?...
$$1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + ... + \frac{1}{1 + 2 + 3 + ... + n}$$

Finding limit of a (Laurent?) series

I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one: 1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + ... + \frac{1}{1 + 2 + 3 + ... + n} The task is to find the limit....

About the set of all the solutions $\mathbf x=(x_1,x_2,\cdots,x_m)$ to $\sum_{j=1}^m\frac{1}{x_j}=\frac1n$

I've just been able to prove that my expectation is true. In order to prove this, let us define the following sequence for n\in\mathbb N : e_{1,n}=n+1,\ \ e_{m,n}=ne_{1,n}e_{2,n}\cdots e_{m-1,n}+1\ \ (m=2,3,4,\cdots). Lemma 1 : begin{align}\frac1{e_{1,n}}+\frac1{e_{2,n}}+\cdots+\frac1{e_{m,n}}=\frac1n-\frac1{e_{m+1,n}-1}.\end{align} Proof for lemma 1 : By the definition, we get (e_{m-1,n}-1)e_{m-1,n}=ne_{1,n}\cdots e_{m-2,n...
$$\frac{x}{n} -1 +\frac{x+1}{n} - 1+\frac{x+2}{n} - 1 \dots +\frac{x+n-1}{n} -1 \lt [\frac{x}{n}]+[\frac{x+1}{n}]+[\frac{x+2}{n}] \dots +[\frac{x+n-1}{n}] \le \frac{x}{n} + \frac{x+1}{n}+\frac{x+2}{n} \dots +\frac{x+n-1}{n}$$

How do I prove that $[\frac{x}{n}]+[\frac{x+1}{n}]+[\frac{x+2}{n}]....+[\frac{x+n-1}{n}]=[x]$

HINT Use the fact that [t] is the only integer such that t - 1 \lt [t] \le t \frac{x}{n} -1 +\frac{x+1}{n} - 1+\frac{x+2}{n} - 1…+\frac{x+n-1}{n} -1 \lt [\frac{x}{n}]+[\frac{x+1}{n}]+[\frac{x+2}{n}]…+[\frac{x+n-1}{n}] \le \frac{x}{n} + \frac{x+1}{n}+\frac{x+2}{n}…+\frac{x+n-1}{n} UPDATE The OP question can be proved using Hermite's identity, by taking x := \frac x n My hint was not useful at all, so please remove the acceptance...
$$x \leq 1+ \frac{1}{2} +\frac{1}{3} + \dots \frac{1}{n-1} + \frac{1}{n} \leq \sum{\frac{1}{m}},$$

Showing the infinitude of primes using the natural logarithm

I came across this proof in Proofs From the Book by Aigner and Ziegler. It uses the inequality logx \leq \pi(x)+1 . (Here, we use natural logarithm) The proof starts with the inequality log x \leq 1+ \frac{1}{2} +\frac{1}{3} + \dots \frac{1}{n-1} + \frac{1}{n} \leq \sum{\frac{1}{m}}, where the sum extends over all m\in \mathbb{N} which have only prime divisors p \leq x My first question is how do we define this sum, do these ...
$$n=k, \\ \dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{k}{1+a_{1}+a_{2}+\cdots+a_{k}} \\ \le\dfrac{k}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}}$$

This general inequality maybe is true? $\sum_{i=1}^{n}\frac{i}{1+a_{1}+\cdots+a_{i}}<\frac{n}{2}\sqrt{\sum_{i=1}^{n}\frac{1}{a_{i}}}$

suppose n=k, \\ \dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{k}{1+a_{1}+a_{2}+\cdots+a_{k}} \\ \le\dfrac{k}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}} when n=k+1 LHS= \dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{k}{1+a_{1}+a_{2}+\cdots+a_{k}}+\dfrac{k+1}{1+a_{1}+a_{2}+\cdots+a_{k+1}} \\<\dfrac{k}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}}+\dfrac{k+1}{1+a_{...
$$ \frac{\frac{1}{2} + x + \dots + x^{n-1} + \frac{x^n}{2}}{n} \leq \frac{x^n+1}{2} $$
$$x_n = 1+ \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$$

Bounded Sequences

I came across the following problems during the course of my self-study of real analysis: Show that the sequence (x_n) defined by x_n = 1+ \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} is unbounded. I know a sequence (x_n) is bounded if there exists a positive number K such that |x_n| \leq K for all n . So suppose for contradiction that it is bounded. Maybe we can define sequences a_n = x_n-1 , b_n = a_n-\frac{1}{2} , ...
$$|\frac{x_1+\dots+x_n}{n}-\frac{1}{2}| \le \epsilon$$

Law of large numbers integrals under continuous functions

You just have the Lebesgue measure on [0,1]^n , and what concentration inequalities from probability tell us is that, f...
$$c_0-1+\frac{1}{10}c_1+...+\frac{1}{10^n}c_n\leq x$$

Courant and Fritz, Constructing the real numbers

In chapter 1, page 10, real numbers are found by confining them to an interval that shrinks to "zero" length (we consider subintervals that approximate better and better the position of x ). Basically, if x is between c and c+1 , then we can divide that interval into ten subintervals, and we can, then, have c+\frac{1}{10}c_1\leq x\leq c+\frac{1}{10}c_1+\frac{1}{10} , where c_1 is a digit from zero to nine. Repeating this proces...
$$c_0-1+\frac{1}{10}c_1+...+\frac{1}{10^n}c_n\leq x$$

Courant and Fritz, Construction of the real numbers | Physics Forums

In chapter 1, page 10, real numbers are found by confining them to an interval that shrinks to "zero" length (we consider subintervals ##I_0,\,I_1,...,\,I_n##). Basically, if ##x## is between ##c## and ##c+1##, then we can divide that interval into ten subintervals, and we can, then, have...
$$\sum \frac{1}{1+x_i} \leq \frac{n-1}{2} + \frac{1}{1+x_1x_2...x_n}$$

An inequality associated with sum of n real numbers

What is the point in defining the y_i ? This is equivalent to show directly that \sum \frac{1}{1+x_i} \leq \frac{n-1}...
$$ \frac{x^n+\dots+x}{n} = x(1-x^n)/(1-x) \leq 0$$

Show that $\sum\limits_{k=1}^{n-1} (n-k) x^k$ is non-decreasing for $x \in ]-1,1[$.

I found it I think. Let's replace n by n+1 to obtain: p(x) = \sum_{k=1}^n(n+1-k)x^k, now we let the sum go from 0 to n-1 : p(x) = \sum_{k=0}^{n-1} (n-k)x^{k+1}, taking the derivative yields: p'(x) = \sum_{k=0}^{n-1}(n-k)(k+1)x^k it now suffices to show that p'(x) \geq 0 on ]-1,1[ . One can check (for example with mathematica) that we have: p'(x) = \frac{n(x^{n+1} + 1) - 2 (x^n + \dots + x))}{(x-1)^2}, t...
$$\left|\frac{a_0+\dots+a_{n-1}x^{n-1}}{x^n}\right|\leq 1$$

Big-O of a polynomial within a logarithm.

We can assume that a_n=1 . Let R such that if |x|>R then \left|\frac{a_0+\dots+a_{n-1}x^{n-1}}{x^n}\right|\leq 1 . Then \log|g(x)|=\log\left(|x|^n\left(\left|\frac{a_0+\dots+a_{n-1}x^{n-1}}{x^n}\right|\right)+1\right)\leq \log(|x|^n+1)\leq \log 2+n\log |x| if |x|>\max\{R,1\} . This gives the wanted result, as if we assume |x|\geq \max\{R,2\} , we get \log|g(x)|\leq (n+1)\log|x| ....
$$1+x\leq\left(1+{x\over n}\right)^{n}\leq{1\over 1-x}.$$

derivative of exponential function

In this entry, we shall compute the derivative of the exponential function from its definition as a limit of powers. If \(0\leq x<1\), then By the inequalities for differences of powers, we have Since \(n-10\), we have \(0<(n-1/n)x1-x\), so Hence Taking the limit as \(n\to\infty\), we obtain our result. ∎ Assume \(0
$$\displaystyle |x-4.5|^{n+1}\le {1\over 2^{n+1}}$$

Find a polynomial which approximates $f(x) = \sqrt{x}$ in the interval $(4,5)$ within $10^{-8}$

By the Lagrange form of the error in Taylor's Remainder Theorem, if f^{(n+1)}(x) is continuous on an open interval I which contains a and x\in I , then there exists some c between a and x such that R_n(x)={f^{(n+1)}(c)\over (n+1)!}(x-a)^{n+1}. My quick calculations yield f(x)=\sqrt{x}\implies f^{(n+1)}(x)={(-1)^n\over 2^{n+1}}(2n-1)!!\,x^{-(2n+1)/2}. Taking a=4.5 (the midpoint of the given interval), we have 4...
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