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anemone
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Determine the sum \(\displaystyle \sum_{k=1}^n \dfrac{4k}{4k^4+1}\).
MarkFL said:Here is my solution:
We are given to evaluate:
\(\displaystyle S_n=\sum_{k=1}^n\left(\frac{4k}{4k^4+1} \right)\)
Partial fraction decomposition on the summand allows us to write:
\(\displaystyle S_n=\sum_{k=1}^n\left(\frac{1}{2k^2-2k+1}-\frac{1}{2k^2+2k+1} \right)\)
Observing that:
\(\displaystyle 2(k+1)^2-2(k+1)+1=2k^2+2k+1\)
and using the rule of linearity of the summand and re-indexing the first sum, we obtain:
\(\displaystyle S_n=\sum_{k=0}^{n-1}\left(\frac{1}{2k^2+2k+1} \right)-\sum_{k=1}^n\left(\frac{1}{2k^2+2k+1} \right)\)
Pulling the first term from the first sum and the last term from the second sum, we may write:
\(\displaystyle S_n=1+\sum_{k=1}^{n-1}\left(\frac{1}{2k^2+2k+1} \right)-\sum_{k=1}^{n-1}\left(\frac{1}{2k^2+2k+1} \right)-\frac{1}{2n^2+2n+1}\)
The two sums add to zero, and we are left with:
\(\displaystyle S_n=1-\frac{1}{2n^2+2n+1}=\frac{2n(n+1)}{2n^2+2n+1}\)
The general formula for finding the sum of the series $\dfrac{4k}{4k^4+1}$ is $\dfrac{1}{2}\ln{\left(\dfrac{k^4+1}{k^4}\right)}$.
Yes, the sum of the series $\dfrac{4k}{4k^4+1}$ can be expressed in a closed form using the natural logarithm function.
The formula for the sum of the series $\dfrac{4k}{4k^4+1}$ can be proved using the technique of partial fractions and telescoping series.
Yes, the formula for the sum of the series $\dfrac{4k}{4k^4+1}$ can be extended to other series with a similar form, such as $\dfrac{ak}{bk^4+1}$ where a and b are constants.
The formula for the sum of the series $\dfrac{4k}{4k^4+1}$ can be used in various fields of science and mathematics, such as in the study of infinite series, calculus, and engineering. It can also be applied in problem-solving and optimization scenarios.