Suppse we are given $$B = \frac{1}{q+1}+\frac{1}{(q+2)(q+1)}+\frac{1}{(q+3)(q+2)(q+1)}+...$$ Then we want to make it to become $$B<\frac{1}{q+1}+\frac{1}{(q+1)^2}+\frac{1}{(q+1)^3}+...$$ Actually why this inequality valids? In other words, how do we know the inequality holds for all terms as we don't know what will happen at larger term.
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Note that $q+a > q+1$ for $a>1$. Hence, we get that $\dfrac1{q+a} < \dfrac1{q+1}$ for $q>-1$. This means for $q>-1$, we have$$\prod_{a=1}^k \dfrac1{q+a} < \prod_{a=1}^k \dfrac1{q+1} = \dfrac1{(q+1)^k} \,\,\,\,\,\,\,\,\, \forall k >1$$ Hence for $q>-1$, we have $$B = \sum_{k=1}^{\infty} \left(\prod_{a=1}^k \dfrac1{q+a} \right) < \sum_{k=1}^{\infty} \dfrac1{(q+1)^k} $$ |
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First you should be more clear about what $q$ can be for this statement to be true. Once you have laid out the restrictions on $q$, compare $B$ to the expression in the inequality term-by-term. Note that for real numbers greater than or equal to $0$, if $a \lt b$ then $\frac{1}{a} \gt \frac{1}{b}$. $q+1 \lt q+n$ where $n \in \mathbb{N} \gt 1$. Don't forget to think about negative $q$ as well. Another tactic: apply induction to $B_n$ with base case $n = 2$. Use $a \lt b$ then $\frac{1}{a} \gt \frac{1}{b}$ and a partial fraction decomposition of the $n+1$ term of the RHS to prove the $n+1$ case. |
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