An integer-valued polynomial is a polynomial with complex coefficients taking values in $\mathbb{Z}$ when all the variables take integer values. For example, $\frac{x^2+3x}{2}$ and $\frac{13x^3+5xy^2}{6}$ are integer-valued polynomials. Clearly, the set of integer-valued polynomials with variables $x_1,x_2,\ldots,x_n$ form a subring of $\mathbb{Q}\left[x_1,x_2,\ldots,x_n\right]$. A result by Polya states that the ring of integer-valued polynomials in one variable $x$ is a free abelian group with basis elements $\binom{x}{k}=\frac{x(x-1)(x-2)\cdots(x-k+1)}{k!}$ for $k=0,1,2,\ldots$.
To answer your question, $x^k$ is an integer-valued polynomial. Therefore, $x^k=\sum_{r=0}^k \,a_r\binom{x}{r}$ for some $a_0,a_1,\ldots,a_n\in\mathbb{Z}$ (obviously, $a_n\neq 0$). Now, $\sum_{m=0}^n\,m^k=\sum_{m=0}^n\,\sum_{r=0}^k\,a_r\binom{m}{r}=\sum_{r=0}^k\,a_k\,\sum_{m=0}^n\,\binom{m}{r}$. By the Hockey-Stick Identity (see http://www.artofproblemsolving.com/wiki/index.php/Combinatorial_identity#Hockey-Stick_Identity), $\sum_{m=0}^n\,m^k=\sum_{r=0}^k\,a_k\,\binom{n+1}{r+1}$. Hence, $\sum_{m=0}^n\,m^k$ is a polynomial in $n$ of degree $k+1$, as the coefficient of $n^{k+1}$ is $\frac{a_k}{(k+1)!}\neq 0$. (In fact, $a_k=k!$, so we know that $\sum_{m=0}^n\,m^k=\frac{n^{k+1}}{k+1}+\mathcal{O}\left(n^k\right)$.)
The alternating series
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}n$$ is well-known to tend to $\log 2$, and the expression inside the main parenthesis oscillates around $1$. One can expect an asymptotic behavior like
$$1\pm\frac1{2n}.$$
Then taking the $n^{th}$ power, the value will alternatively rejoin $e^{1/2}$ and $e^{-1/2}$, so the limit of the sequence does not exist.
More precisely, if we group the terms in pairs, we have alternatively
$$S_{2n}=1+\sum_{k=2n+2}^\infty\frac1{2k(2k+1)}\sim 1+\frac1{4n}$$
and
$$S_{2n+1}=1+\sum_{k=2n+2}^\infty\frac1{2k(2k+1)}-\frac1{2n+1}\sim 1-\frac1{4n},$$ approximating the sums by integrals.
Taking the power, we have
$$S_{2n[+1]}^{2n}\sim\left(1\pm\frac1{4n}\right)^{2n}\sim e^{\pm1/2}.$$
Best Answer
You should add what you've tried. Anyway:
Use induction for the 1st argument. Obviously $S_1=1$ and $S_2=2=1/2+(1+1/2)S_1$, so the proposition is true for $n=2$. Assume that it is true for some $n$ and with that in hand prove it for $n+1$: We have $S_{n+1}=$ terms that $n+1$ does not appear on the denominator $+$ terms that $n+1%$ appears on the denominator $=S_n+\frac{1}{n+1}S_n+\frac{1}{n+1}$, since $\frac{1}{n+1}$ is a common factor of the elements that have $(n+1)$ as a divisor of the denumerator. By induction, this formula is true for all $n\in\mathbb{N}$. For the 2nd argument, use induction again. We already saw that $S_1=1$. Assume that $S_{n}=n$. Then use the formula proved above and get $S_{n+1}=n+\frac{n+1}{n+1}=n+1.$ These things are quite simple, try hard and don't give up easily.