Infinite Sum
- Tolaso J Kos
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Infinite Sum
If \( a \in \mathbb{N} \) show that: $$\sum_{k=1}^{\infty}\frac{k}{k^4+4a^4}=\frac{1}{4a}\sum_{k=1}^{2a}\frac{1}{\left ( k-a \right )^2+a^2}$$
Imagination is much more important than knowledge.
- Tolaso J Kos
- Administrator
- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
- Contact:
Re: Infinite Sum
We see that the sum is telescopic because:
$$\begin{aligned}
\frac{k}{k^4+4a^4} &=\frac{k}{\left ( k^2+2a^2 \right )^2-4k^2a^2} \\
&= \frac{k}{\left ( k^2-2ka+2a^2 \right )\left ( k^2+2ka+2a^2 \right )}\\
&=\frac{1}{4a}\frac{\left ( k^2+2ka+2a^2 \right )-\left ( k^2-2ka+2a^2 \right )}{\left ( k^2-2ka+2a^2 \right )\left ( k^2+2ka+2a^2 \right )} \\
&=\frac{1}{4a}\left [ \frac{1}{k^2-2ka+2a^2}-\frac{1}{k^2+2ka+2a^2} \right ]
\end{aligned}$$
Thus for \( a \in \mathbb{N} \) our desired sum is equal to:
$$\sum_{k=1}^{\infty}\frac{k}{k^4+4a^4} =\frac{1}{4a}\sum_{k=1}^{\infty}\left [ \frac{1}{k^2-2ka+2a^2}-\frac{1}{k^2+2ka+2a^2} \right ] =\frac{1}{4a}\sum_{k=1}^{2a}\frac{1}{\left ( k-a \right )^2+a^2}$$
$$\begin{aligned}
\frac{k}{k^4+4a^4} &=\frac{k}{\left ( k^2+2a^2 \right )^2-4k^2a^2} \\
&= \frac{k}{\left ( k^2-2ka+2a^2 \right )\left ( k^2+2ka+2a^2 \right )}\\
&=\frac{1}{4a}\frac{\left ( k^2+2ka+2a^2 \right )-\left ( k^2-2ka+2a^2 \right )}{\left ( k^2-2ka+2a^2 \right )\left ( k^2+2ka+2a^2 \right )} \\
&=\frac{1}{4a}\left [ \frac{1}{k^2-2ka+2a^2}-\frac{1}{k^2+2ka+2a^2} \right ]
\end{aligned}$$
Thus for \( a \in \mathbb{N} \) our desired sum is equal to:
$$\sum_{k=1}^{\infty}\frac{k}{k^4+4a^4} =\frac{1}{4a}\sum_{k=1}^{\infty}\left [ \frac{1}{k^2-2ka+2a^2}-\frac{1}{k^2+2ka+2a^2} \right ] =\frac{1}{4a}\sum_{k=1}^{2a}\frac{1}{\left ( k-a \right )^2+a^2}$$
Imagination is much more important than knowledge.
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