Tagged Questions

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Prove natural log between two finite harmonic sums [duplicate]

Prove for n in the naturals we have: $$\sum_{k=2}^n 1/k \le \ln(n) \le \sum_{k=1}^{n-1} 1/k$$ Intuitively this makes sense to me but I can't for the life of me figure out how to start this proof.
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How to show $\log_2 x \cdot \log_{0.25} x \cdot \log_{0.125} x \cdot \log_{16} x > \frac {2}3$?

I was trying to solve $\log_2 x \cdot \log_{0.25} x \cdot \log_{0.125} x \cdot \log_{16} x > \frac {2}3$ and I keep getting a partial answer of $x>4$ though answer key suggests a more expanded ...
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Is $ln(x)$ ever greater than $x$

Is $\forall x \in \mathbb{R}, \ln(x) \lt x$ a true statement? Just wondering for some convergence related thing
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$\ln(n)/n<1/2$ proof without calculus or any kind of advanced mathematics

Is it possible to show that $\ln(n)/n<1/2$, for all natural numbers $n$ without using calculus, but just some elementary math? Induction is allowed. I was trying to show equivalently that ...
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Tighter logarithmic inequality

There is a well-known lower bound for $$x\log{1+x\over x}\geq {x\over1+x}$$ for $x\geq0$. I know a tighter lower bound on the same domain $$x\log{1+x\over x}\geq{2x\over1+2x}\geq {x\over1+x}.$$ It ...
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Something about $\frac{\log x}{x}$

Denote $\log x = \log_ex$. Let's consider the below function $$\frac{\log x}{x}$$. Apparently, It's maximum is $\frac{1}{e}$. and strictly increasing in $(0,e]$, strictly decreasing in $[e,+\infty)$. ...
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Upperbound for $\sum _{i=1}^N a_i\ln a_i$

It's easy to prove that following upperbound is true: $\sum_{i=1}^N a_i\ln a_i \le A \ln A$, where $\sum_{i=1}^N a_i=A$ and $a_i\ge 1$ I'm wondering, is there stronger upperbound?
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Why does this equation work?

let $P(x) := \sum_{p \leq x} Log [p]$, then we have $P(2^{k+1}) = \sum_{i=0}^k ( P(2^{i+1}) - P(2^i)) < 2 \cdot Log[2] \cdot (1 + 2 + 4 +... + 2^k) \leq 4 \cdot Log[2] \cdot 2^k$. Why does ...
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Logarithmic equation

I'm studying logarithms and I encountered this equation: $$[\log_9(k+1)]^2+\log_9(k+1)+(k+1)>3$$ I tried a lot but I still couldn't solve it! I know this may be easy for most of you but please ...
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Parametric inequation…

Supppose we have $a$ a real positive number that's not equal to $1$. Solve the following inequation: $$\log_a(x^2-3x)>\log_a(4x-x^2)$$ If it's known that $x=3.75$ is one solution of it.
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Help with integral/logarithm inequality

I have to prove the following inequality: $1/(n+1) < \int_n^{n+1} 1/t$ $dt$ $<1/n$ I thought it would be easier to attack this via integration, so I get: $1/(n+1) <$ log $(n+1)-$ ...
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Solving inequality having log

I am struggling to solve this inequality involving logarithm. How to find out values of $n$ for which below inequality holds good: $${\log_2n \over n} >{ 1 \over 8}$$
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Inequality $C\lceil\log{n}\rceil! \geq n^k$

I've been struggling to prove there exist $C$ for $n, n_{0}, \forall k >0 \in \mathbb{R}$ such that $\forall n > n_{0}$: $$C\lceil\log{n}\rceil! \geq n^k$$ As you ...
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Inequality with logarithms

How do I show that $$\frac{1}{n-1}\geq \ln \left ( \frac{n}{n-1} \right )$$ for $n>1$? As far as I can tell, exponentiating both sides with base $e$ won't help, because then I get a nasty ...
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determine x in $x\log_\frac{1}{10}(x^2+x+1)>0$

I wanted to know, how can i determine the values of x for which $x\log_\frac{1}{10}(x^2+x+1)>0$ going to the question, we must have $x>0$ and $\log_\frac{1}{10}(x^2+x+1)>0$ or both must ...
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If $ax^2-bx+c=0$ has two distinct real roots lying in the interval $(0,1)$ $a,b,c$ belongs to natural prove that $\log_5 {abc}\geq2$

If $ax^2-bx+c=0$ has two distinct real roots lying in the interval $(0,1)$ with $a, b, c\in \mathbb N$, prove that $\log_5 {abc}\geq2$. The equations I could form are: 1) $f(0)>0$ and ...
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Do these inequalities regarding the gamma function and factorials work?

I am seeing if it is possible to generalize lemma 1 in Nagura's proof that there is always a prime between $x$ and $\frac{6x}{5}$. In a previous question, I asked whether the following inequality is ...
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Trying to generalize an inequality from Jitsuro Nagura: Does this work?

I am investigating the generality of Lemma 1 in Nagura's proof that there is always a prime between $x$ and $\frac{6x}{5}$. In Lemma 1, Nagura establishes when $n > 1$, $x \ge 1$: ...
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For $x > 1$, $n > 2$ with $2 \mid x+1$ and $n \mid x+1$, does it then follow that: $$\log(\lfloor\frac{x}{2}\rfloor!) - \log(\lfloor\frac{x}{n}\rfloor!) \le \log(\lfloor\frac{x+1}{2}\rfloor!) - ... 4answers 2k views Prove \left(\dfrac{2}{5}\right)^{\frac{2}{5}}<\ln{2} Inadvertently, I find this interesting inequality,But this problem have nice solution? prove that$$\ln{2}>(\dfrac{2}{5})^{\frac{2}{5}}$$This problem have nice solution? Thank you. ago,I find ... 1answer 120 views Understanding a famous proof by Jitsuro Nagura: Need help understanding one step in the main theorem I am going through the proof by Jitsuro Nagura which shows that there is always a prime between x and \frac{6x}{5} where x \ge 25. Nagura uses the following definitions:$$\vartheta(x) = ...
I need to find sufficient conditions so that $x \geq \frac{1}{a-\ln{x}}$ for $a>1$ and $x > 0$. Is there a way to get a tight solution to the problem?