Theorem : If |r|<1|r| \lt 1 and p∈Z+p \in \mathbb Z^+ then nprnn^pr^n converges to 00 .
We want to prove this theorem in some specific steps.
Show that its enough to consider the case in which 0

Select a positive number like aa such that 1+a=1r1+a=\frac{1}{r} and explain that why ∀n>2p+2\forall n\gt2p+2 we have

1rn>np+12p+1(p+1)!ap+1\frac{1}{r^n} \gt \frac{n^{p+1}}{2^{p+1}(p+1)!}a^{p+1} .

Prove that for each large enough nn, nprn<2p+1(p+1)!nap+1n^pr^n \lt \frac{2^{p+1}(p+1)!}{na^{p+1}} and complete the proof of the theorem.
I have no idea how to prove steps 22 to 44.
Any hint would be appreciated.
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Note that (nk)=n(n−1)⋅…⋅(n−k+1)k!\binom{n}{k} = \frac{n(n-1)\cdot\dotsc \cdot (n-k+1)}{k!}. Use the binomial formula. Combine.
– Daniel Fischer♦
2 days ago
@DanielFischer Which part of the proof are u talking about ? 🙂
– IStillHaveHope
2 days ago
Steps 2, 3, and 4. But I made a mistake for step 3 initially.
– Daniel Fischer♦
2 days ago
@DanielFischer well, i thought but i didn't get what u meant ... what should i combine ?
– IStillHaveHope
2 days ago
@DanielFischer can you please explain your idea? 🙂
– IStillHaveHope
2 days ago
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2 Answers
2
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For step 2, use
(nk)=n!k!(n−k)!=n⋅(n−1)⋅…⋅(n−k+1)k!\binom{n}{k} = \frac{n!}{k!(n-k)!} = \frac{n\cdot (n-1)\cdot \dotsc \cdot (n-k+1)}{k!}
and note that for n⩾2kn \geqslant 2k, each of the kk factors in the numerator is > \frac{n}{2}> \frac{n}{2}.

In step 3, by the binomial formula

\frac{1}{r^n} = (1+a)^n = \sum_{k = 0}^n \binom{n}{k} a^k \geqslant \binom{n}{m} a^m\frac{1}{r^n} = (1+a)^n = \sum_{k = 0}^n \binom{n}{k} a^k \geqslant \binom{n}{m} a^m

for every fixed 0 \leqslant m \leqslant n0 \leqslant m \leqslant n, since a > 0a > 0. Choose an appropriate mm, and combine the resulting inequality with the result of step 2. Finally, cross-multiply to arrive at the inequality of step 4. Use the fact that \frac{2^{p+1}(p+1)!}{a^{p+1}}\frac{2^{p+1}(p+1)!}{a^{p+1}} is a constant independent of nn to reach the conclusion.

2) Write out \binom{n}{k}\binom{n}{k} and multiply by 2^n k!2^n k! and (n-k)!(n-k)! to get:

2^n n\cdot(n-1)\cdot…\cdot(n-k+1) > n^k2^n n\cdot(n-1)\cdot…\cdot(n-k+1) > n^k

and 2^n n\cdot(n-1)\cdot…\cdot(n-k+1) = 2^n \prod^k_{j=1}(n-k+j) > \prod^k_{j=1}2(n-k+j)2^n n\cdot(n-1)\cdot…\cdot(n-k+1) = 2^n \prod^k_{j=1}(n-k+j) > \prod^k_{j=1}2(n-k+j)

Now since n>2kn>2k every factor is bigger than nn and therefore the expression is bigger than n^kn^k.