Proofs involving summations

Question

I am working on the following proof.

Prove that $\displaystyle \sum_{k=1}^{n} \frac{2k^4+1}{k+2} \geq \frac{n(n+1)(2n+1)}{6}$.

My first question is if the above is equivalent to saying "Prove that:$\displaystyle \sum_{k=1}^{n} \frac{2k^4+1}{k+2} \geq \displaystyle \sum_{n=1}^{t} \frac{n(n+1)(2n+1)}{6}$." If this was in fact correct, I examined the partial sums and found

$1+\frac{33}{4}+\frac{163}{5}+\frac{171}{2}+ \cdots \geq 1+2^2+3^2+4^2+ \cdots$. From here I see that when $n=1$ the inequality turns into an equality. My plan from here was to use some kind of induction to show that $\forall k_{i}, i,j \in \mathbb{N}_, k_{i} \geq n_{j}$.

Other than this method I do not see another way to go, could someone offer a hint of a different route or some help with my current route.

For my current route I am confused on the bounds for the second summation $\displaystyle \sum_{n=1}^{t} \frac{n(n+1)(2n+1)}{6}$.

Thank you.

Answer 1

$$\text {We have }\;\frac {n(n+1)(2n+1)}{6}=\sum_{k=1}^n k^2.$$ $$ \text {Therefore }\;\sum_{k=1}^n \frac {2k^4+1}{k+2}-\frac {n(n+1)(2n+1)}{6}=\sum_{k=1}^n A_k$$ where $$A_k=\frac {2k^4+1}{k+2}-k^2=\frac {[2k^4-2k^2] -[k^3-1]}{k+2)}=\frac {k-1}{k+2}([2k^3-k] +[k^2-1])\geq 0.$$

Answer 2

Presumably your equivalent statement should read $$ \sum_{k=1}^n\frac{2k^4+1}{k+2}\geq \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$$

If so, then yes this is true, since $\frac{2k^4+1}{k+2}=1=k^2$ when $k=1$, and for $k\geq 2$ we have $$ \frac{2k^4+1}{k+2}\geq \frac{2k^4+1}{k+k}\geq \frac{2k^4}{2k}=k^3\geq k^2$$

Answer 3

There is no reason to sum the second expression in the equality; all that is needed is a series of inequalities $$\displaystyle \sum_{k=1}^{n} \frac{2k^4+1}{k+2} > \sum_{k=1}^n \frac{2k^4}{2k} = \sum_{k=1}^n k^3 > \sum_{k=1}^n k^2 =\frac{n(n+1)(2n+1)}{6}$$

If you know that summing an $n$th degree polynomial yields a polynomial of degree $n+1$ then the answer is simple; you start by summing a polynomial $O(n^3)$ which will yield a fourth degree polynomial $O(n^4)$. The fraction on the right is $O(n^3)$, and so we are done.

Answer 4

I think what you are asking is $\frac{2k^4 + 1}{k + 2} \ge \frac{k(k+1)(2k+1)}{6}$.

$\frac{k(k+1)(2k+1)}{6}= \frac{k(k+1)(k+2)(2k+1)}{6(k+2)}=\frac{\frac 16(2k^4 + 7k^3 +7k^2 + 2k)}{k+2}$

So your question is true if and only if $2k^4 + 1 \ge \frac 16(2k^4 + 7k^3 +7k^2 + 2k)$

Which is true if and only if $ 10k^4 + 6 \ge 7k^3 + 7k^2 + 2k$

Which is true if and only if $10k + 6/k^4 \ge 7 + 7/k + 2/k^2$.

Which is true for $k = 1$ and as the LHS increases and the RHS decreases, this is always true.

Which is something I think the authors of the problem didn't notice.

So $\sum_{k=1}^{n}\frac{2k^4 + 1}{k + 2} = \sum_{k=1}^{n-1}\frac{2k^4 + 1}{k + 2}+ \frac{2n^4 + 1}{n+2}$

$\ge 0 + \frac{2n^4 + 1}{n+2} = \frac{2n^4 + 1}{n+2}$

$\ge \frac{n(n+1)(2n+1)}{6}$