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Let $a_1=1, a_{n+1}=a_n^2+a_n(n \geq 2)$. Prove

$$\sum_{k=1}^n \frac{2}{2a_k+1} \leq \frac{22}{15}-\frac{1}{\left(\frac{3}{2}\right)^{2^n}-1}.$$

I've proven a lemma that $$\forall n:\left(a_n+\frac{1}{2}\right)^2\geq2^{n}$$ by induction. This will help?

Maybe we can obtain $$\forall n:a_n+\frac{1}{2}\geq \left(\frac{3}{2}\right)^{2^{n-1}}$$

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  • $\begingroup$ Do you want to prove it by induction? $\endgroup$ – Dr. Sonnhard Graubner Nov 10 '19 at 12:37
  • $\begingroup$ @ you'd better not. $\endgroup$ – mengdie1982 Nov 10 '19 at 12:38
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    $\begingroup$ Your lemma is unnecessarily weak. It's not hard to show $$a_n + \frac{1}{2} \geqslant \Bigl(\frac{3}{2}\Bigr)^{2^{n-1}}.$$ $\endgroup$ – Daniel Fischer Nov 10 '19 at 13:57
  • $\begingroup$ @DanielFischer Can you give a proof based on your result? $\endgroup$ – mengdie1982 Nov 10 '19 at 13:59
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The key here is that the sequence is rapidly increasing so that $2a_n+1\approx2a_n$ and thus for "large" $n$ we have

$${2\over2a_n+1}\approx{1\over a_n}={1\over a_{n-1}(a_{n-1}+1)}={1\over a_{n-1}}-{1\over a_{n-1}+1}$$

The trick is to figure out (or guess) how many terms to keep exact, and then how to turn the approximation into a useful inequality.

Now the sequence $a_1,a_2,a_3,\ldots$ is $1,2,6,42,\ldots$, so

$$\sum_{k=1}^n{2\over2a_k+1}={2\over3}+{2\over5}+{2\over13}+{2\over86}+\cdots={22-6\over15}+{2\over13}+{2\over86}+\cdots$$

so, after checking the cases $n=1$ and $2$, the inequality is equivalent to

$${1\over(3/2)^{2^n}-1}+\left({2\over13}+{2\over86}+\cdots\right)\le{2\over5}$$

for $n\ge3$. But

$$\begin{align} {2\over13}+{2\over86}+\cdots &={2\over2a_3+1}+{2\over2a_4+1}+{2\over2a_5+1}+\cdots\\ &\le{1\over a_3}+{1\over a_4}+{1\over a_5}+\cdots\\ &={1\over a_2(a_2+1)}+{1\over a_3(a_3+1)}+{1\over a_4(a_4+1)}+\cdots\\ &={1\over a_2}-{1\over a_2+1}+{1\over a_3}-{1\over a_3+1}+{1\over a_4}-{1\over a_4+1}+\cdots\\ &={1\over2}-{1\over3}+{1\over6}-{1\over7}+{1\over42}-{1\over43}+\cdots\\ &\le{1\over2}-{1\over3}+{1\over6}\\ &={1\over3} \end{align}$$

so it suffices to check that

$${1\over(3/2)^{2^n}-1}\le{2\over5}-{1\over3}={1\over15}$$

for $n\ge3$. Since $(3/2)^{2^n}$ is increasing, it's enough to verify $1/((3/2)^8-1)\le1/15$, which is equivalent to $16\le(3/2)^8$, which is easy to see: $(3/2)^4=81/16\gt5$, so $(3/2)^8\gt5^2=25\gt16$.

Just for completeness, let's actually check the cases $n=1$ and $2$:

$${22\over15}-{1\over(3/2)^2-1}={22\over15}-{4\over9-4}={22\over15}-{4\over5}={22-12\over15}={2\over3}$$

and

$${22\over15}-{1\over(3/2)^4-1}={22\over15}-{16\over81-16}={22\over15}-{16\over65}\gt{22\over15}-{26\over65}={22\over15}-{2\over5}={16\over15}={2\over3}+{2\over5}$$

(so the inequality is strict except for $n=1$).

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