Implicit definition of recursive sequence $\displaystyle a_{n+1} = a_n + \frac{1}{(3 + (-1)^n)^n}$
By Sophia Hammond
I have to show that the recursive sequence given by $$ b_1 = 1, ~ b_{n+1} = b_n + \frac{1}{(3 + (-1)^n)^n} $$ converges.
I can show its convergence by showing its monotone and bounded by 2 but I wondered, if I can find a implicit definition of the sequence without an $(-1)^n$ terms in it. Obviously, $$b_n = b_1 + \sum_{k = 0}^{n-1} \frac{1}{(3 + (-1)^n)^n}$$
I figured out that
$$ b_{2n} = \sum_{k = 0}^{n} \frac{1}{16^k} + \frac{1}{2 \cdot 4^k} = \sum_{k = 0}^{n} \frac{4^k + 2}{2^{4k+1}} $$
which converges by the geometric series test (also WolframAlpha says so), whilst it WolframAlpha doesn't make any statement about the convergence of $\displaystyle \sum_{k = 0}^{n-1} \frac{1}{(3 + (-1)^n)^n}$.
Is there a implicit definition of $(b_n)_{n \in \mathbb{N}}$ without the $(-1)^n$, or is that simply not possible?
$\endgroup$ 43 Answers
$\begingroup$The sequence is increasing. It sufficient to show that it is bounded.
We have $$b_{n+1}-b_n=\frac 1{(3+(-1)^n)^n}\leqslant\frac1{2^n}\implies b_n-b_1\leqslant \frac1{2^{n-1}}+\frac1{2^{n-2}}+\ldots+\frac1{2}<1$$
$\endgroup$ $\begingroup$It is easy to find a formula for $b_n$ separating the cases in which $n$ es even or odd. But you do not need this to prove convergence of the series, since $$ 0\le\frac 1{(3+(-1)^n)^n}\le\frac1{2^n}. $$
$\endgroup$ $\begingroup$Observe that $$ a_{k+1}-a_1=\sum_{n=1}^k\frac{1}{(3+(-1)^n)^n}\tag{1} $$ But $$ 0\leq\sum_{n=1}^\infty\frac{1}{(3+(-1)^n)^n}=L<\infty $$ since $$ \frac{1}{4^n}\leq\frac{1}{(3+(-1)^n)}\leq\frac{1}{2^n} $$ for $n\ge1$. let $k\to\infty$ in (1) to deduce that $$ \lim_{k\to\infty}a_{k+1}=\lim_{k\to\infty}(a_{k+1}-a_{1})+\lim_{k\to\ \infty}a_{1} $$ as desired.
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