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Problem 3. Discuss convergence of $\alpha _n := \frac{1}{n^5}\sum_{k=1}^n k^4 $.

[Solution] (1) We know that 
\begin{align}
x^5 - y^5 = (x-y)(x^4 +x^3 y + x^2 y^2 + x y ^3 +y^4)\notag
\end{align}
Then
\begin{align}
2^5 - 1^5 &= \quad 2^4 + 2^3\cdot 1 + 2^2\cdot 1^2 + 2\cdot 1^3 + 1^4 \notag\\
3^5 - 2^5 &= \quad 3^4 + 3^3\cdot 2 + 3^2\cdot 2^2 + 3\cdot 2^3 + 2^4 \notag\\
&\vdots\notag\\
(n+1)^5 - n^5 &= \quad (n+1)^4 + (n+1)^3\cdot n + (n+1)^2\cdot n^2 + (n+1)\cdot n^3 + n^4 \notag
\end{align}
From all above, we find that
\begin{align}
(n+1)^5 - 1 &= \sum (k+1)^4 + \sum (k+1)^3 k + \sum (k+1)^2 k^2 + \sum (k+1) k^3 + \sum k^4\notag\\
&= 5\sum k^4 + 10 \sum k^3 + 10 \sum k^2 + 5 \sum k + n\notag
\end{align}
After simplification, we get
\begin{align}
\sum_{k=1}^n k^4 = \frac{1}{5}n ^5 + \frac{1}{2} n ^4 + \frac{1}{3} n ^3 - \frac{1}{30}n.\notag
\end{align}

To show $(\alpha_n)$'s convergence. We guess the limit is $\frac{1}{5}$. Estimating the error
\begin{align}
|\alpha_n -\frac{1}{5}| = \frac{1}{n} (\frac{1}{2} + \frac{1}{3n} - \frac{1}{30n^3}) \textless \frac{1}{n}\cdot (\frac{1}{2} + \frac{1}{3} + 0 ) = \frac{5}{6} \frac{1}{n}.\notag
\end{align}
Given $\varepsilon\textgreater 0 $, choose $N_\varepsilon = [\frac{5}{6\varepsilon}]+1$. Then when $n\geq N_\varepsilon$,
\begin{align}
|\alpha_n - \frac{1}{5}| \textless \frac{5}{6} \frac{1}{n} \leq \frac{5}{6}\frac{1}{N_\varepsilon} \leq \frac{5}{6}\frac{1}{  \frac{5}{6\varepsilon} } = \varepsilon\notag
\end{align}
Hence $(\alpha_n)$ converges to $\frac{1}{5}$. 



[Note] I prefer another way for computation. 
\begin{align}
(1+1)^5 &= 1^5 + 5\cdot 1^4 + 10\cdot 1^3 + 10\cdot 1^2 + 5\cdot 1 + 1\notag\\
(2+1)^5&= 2^5 + 5\cdot 2^4 + 10\cdot 2^3 + 10\cdot 2^2 + 5\cdot 2 + 1\notag\\
&\vdots\notag\\
(n+1)^5&= n^5 + 5\cdot n^4 + 10\cdot n^3 + 10\cdot n^2 + 5\cdot n + 1\notag
\end{align}
Then summing them, 
\begin{align}
(n+1)^5  = 1 + 5\sum k^4 + 10 \sum k^3 + 10 \sum k^2 + 5 \sum k + n\notag
\end{align}
and we will get the same sum.


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Problem 4. Does $g_n:=\frac{1}{n}\sum_{k=1}^n (-1)^{k-1} k$ converge?

[Solution] We compute some terms:
\begin{align}
1,-\frac{1}{2},\frac{2}{3},-\frac{1}{2},\frac{3}{5},-\frac{1}{2},...\notag
\end{align}
i.e.
\begin{align}
g_{2k-1}&=\frac{k}{2k-1}\notag\\
g_{2k}&=-\frac{1}{2}.\notag
\end{align}
But we find that $(g_{2k-1})_{k=1}^\infty$ goes to $\frac{1}{2}$ while $(g_{2k})_{k=1}^\infty $ to $-\frac{1}{2}$. Hence $(g_n)$ diverges. 

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[Note] There is a corollary saying that any subsequence of a convergent sequence must converge to the limit as the original sequence do. This indicates a good way to test a divergent sequence: to find two subsequences converging to different limits. 


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