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Let $n\ge 3$. How to prove show that $$\dfrac{n}{3^n-2^n}<\left(\dfrac{2}{5}\right)^{n-1}\tag1$$ or $$3^n\cdot 2^{n-1}-2^{2n-1}-5^{n-1}\cdot n>0,n\ge 3\tag2$$ It seem right:see wolfram when I do this problem found it:How prove this $\sum_{k=1}^{n}\frac{k}{3^k-2^k}<\frac{5}{3}$

if this inequality (1) has prove it,then $$\sum_{k=1}^{n}\dfrac{k}{3^k-2^k}<1+\dfrac{2}{5}+\sum_{k=3}^{+\infty}\dfrac{2^{k-1}}{5^{k-1}}=\dfrac{5}{3}$$

Question: How to prove inequality (1) or (2)

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  • $\begingroup$ $$\frac{3}{19}<\left(\frac{2}{5}\right)^3$$ is false wolframalpha.com/input/?i=3%2F(27-8)%3C(2%2F5)%5E3 $\endgroup$ – Raffaele Nov 3 '17 at 8:29
  • $\begingroup$ Check that $2^{n+1}$ again. $\endgroup$ – Gerry Myerson Nov 3 '17 at 8:35
  • $\begingroup$ @GerryMyerson,@Raffaele and Hello,I have edit it.Thanks $\endgroup$ – function sug Nov 3 '17 at 8:39
  • $\begingroup$ You can only ping one person per comment. $\endgroup$ – Gerry Myerson Nov 3 '17 at 8:43
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We see that $(1)$ is equivalent to $$6^{n}\gt 4^{n}+2n\cdot 5^{n-1}\tag3$$ Let us prove by induction that $(3)$ holds for $n\ge 3$.

We see that $(3)$ holds for $n=3,4,5$ :

  • $6^3=216\gt 214=4^3+2\cdot 3\cdot 5^{3-1}$

  • $6^4=1296\gt 1256=4^4+2\cdot 4\cdot 5^{4-1}$

  • $6^5=7776\gt 7274=4^5+2\cdot 5\cdot 5^{5-1}$

Supposing that $(3)$ holds for some $n\ (\ge 5)$ gives $$\begin{align}6^{n+1}&=6\cdot 6^n\\\\&\gt 6(4^{n}+2n\cdot 5^{n-1})\\\\&=6\cdot 4^n+12n\cdot 5^{n-1}\\\\&\ge 4\cdot 4^n+(10n+10)\cdot 5^{n-1}\\\\&=4^{n+1}+2(n+1)\cdot 5^n\qquad\blacksquare\end{align}$$

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Some thoughts-not a full answer!

Let's try induction. For $n=3$ we readily verify that it holds: $$\frac{3}{3^3-2^3}\lt\Big(\frac25\Big)^2\Rightarrow\frac3{19}\lt\frac4{25}\Rightarrow75\lt76$$

We assume that the statement is true for $n$. We wish to show that $$\dfrac{n+1}{3^{n+1}-2^{n+1}}<\left(\dfrac{2}{5}\right)^{n}$$

But we know that $$\dfrac{n}{3^n-2^n}\lt\left(\dfrac{2}{5}\right)^{n-1}$$

We multiply this with $\Large\frac25$ to obtain $\Big(\dfrac25\Big)\dfrac{n}{3^n-2^n}\lt\left(\dfrac{2}{5}\right)^{n}$.

Thus it suffices to show that $$\Big(\dfrac25\Big)\dfrac{n}{3^n-2^n}\gt\dfrac{n+1}{3^{n+1}-2^{n+1}}\Rightarrow\\ \\ $$ $$ \Big(\dfrac25\Big)\dfrac{3^{n+1}-2^{n+1}}{3^{n}-2^{n}}\gt\frac{n+1}{n}\Rightarrow\\ $$ $$\large\Big(\dfrac25\Big)\dfrac{\frac{3^{n+1}}{3^{n+1}}-(\frac23)^{n+1}}{\frac{3^{n}}{3^{n+1}}-\frac{2^n}{3^{n+1}}}\gt1+\frac1n\Rightarrow\\ $$ $$\large\Big(\dfrac25\Big)\dfrac{1-(\frac23)^{n+1}}{\frac{1}{3}-\frac13(\frac{2}{3})^n}\gt1+\frac1n\Rightarrow\\$$

$$ \Big(\dfrac65\Big)\dfrac{{1-(\frac23)^{n+1}}}{1-(\frac23)^n}\gt1+\frac1n$$

Now if we show that the function $$f(x)=\Big(\dfrac65\Big)\dfrac{{1-(\frac23)^{x+1}}}{1-(\frac23)^x}-1-\frac1x, x\in[1,+\infty)$$ is strictly increasing we are done since $f(1)=0$ and we will have $f(x)\gt f(1)=0$.

This appears to be the case-here is a graph illustrating it enter image description here

Unfortunately I can't seem to go far when trying to prove that $f$ is strictly increasing.

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$(2)$ can be proved by induction. $(2)$ is equivalent to $$3^n\cdot 2^{n-1}>2^{2n-1}+5^{n-1}\cdot n\quad \quad (3)$$

It is easy to check the cases $n=3,4,5,6$.

Suppose $(3)$ is true for $n=k$. Then $$3^{k+1}\cdot 2^k = 6(3^k\cdot 2^{k-1})>6(2^{2k-1}+5^{k-1}\cdot k)=\frac{3}{2}\cdot 2^{2k+1}+\frac{6}{5}\cdot 5^{k}\cdot k > 2^{2k+1}+\frac{6}{5}\cdot 5^{k}\cdot k$$ Provided $k>5$, $\frac{6}{5}k>k+1$, so the statement holds for all $n>5$ by induction.

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$(A)\enspace$ Induction

$n=3$ : $\enspace$ That’s clear (or see $(B)$) .

Be $\enspace\displaystyle \left(\frac{6}{5}\right)^n - \left(\frac{4}{5}\right)^n - \frac{2n}{5} > 0 \enspace$ (equivalent to $(1)$)$\enspace$ for $\enspace n\geq 4\,$ .

It follows:

$\displaystyle \left(\frac{6}{5}\right)^{n+1} - \left(\frac{4}{5}\right)^{n+1} - \frac{2(n+1)}{5} = $

$\displaystyle =\left(\frac{6}{5}\right)^n - \left(\frac{4}{5}\right)^n - \frac{2n}{5} + \frac{1}{5}\left(\left(\frac{6}{5}\right)^n + \left(\frac{4}{5}\right)^n -2\right) $

$\displaystyle > \frac{1}{5}\left(\left(\frac{6}{5}\right)^4 -2\right) = \frac{1}{5}(2.0736 - 2) > 0 $


$(B)\enspace$ Derivation

$x=3$ : $\enspace\displaystyle \frac{3}{19}=\frac{x}{3^x-2^x}|_{x=3}<\left(\frac{2}{5}\right)|_{x=3}=0.16$

$x=4$ : $\enspace\displaystyle \frac{4}{65}=\frac{x}{3^x-2^x}|_{x=4}<\left(\frac{2}{5}\right)|_{x=4}=0.064$

$x\in\mathbb{R}$ with $x\geq 5$ , $(1)$ multiplicated by $(3^x-2^x)\frac{2}{5}$ :

$\enspace\enspace\displaystyle \frac{2x}{5} < \left(\frac{6}{5} \right)^x - \left(\frac{4}{5}\right)^5 \leq \left(\frac{6}{5} \right)^x - \left(\frac{4}{5}\right)^x $

The right side is clear, the left side we proof by derivation.

$x=5$ : $\enspace\displaystyle 2 = \frac{2x}{5}|_{x=5} < \left(\frac{6}{5}\right)^x|_{x=5} - \left(\frac{4}{5}\right)^5 = 2.16064 $

It’s

$\displaystyle \frac{2}{5} = \frac{d}{dx} \frac{2x}{5} < \frac{d}{dx} (\left(\frac{6}{5}\right)^x - \left(\frac{4}{5}\right)^5 ) = \left(\frac{6}{5}\right)^x \ln \frac{6}{5} $

because of

$\displaystyle \left(\frac{6}{5}\right)^x \geq \left(\frac{6}{5}\right)^5 = 2.48832 > \frac{2}{5} / \ln \frac{6}{5} \approx 2.1939… \enspace$ .

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