Summation Challenge #1: Evaluate $\sum$

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SUMMARY

The summation challenge evaluates the infinite series $$\sum_{k=1}^{\infty} (-1)^{\left\lfloor \frac{k+3}{2} \right\rfloor} \frac{1}{k}$$, which converges to the value $$\frac{\pi}{4} + \frac{1}{2}\ln(2)$$. The series is broken down into two components: an alternating series of odd-indexed terms and a scaled alternating series of even-indexed terms. The analysis demonstrates the application of series convergence techniques and the use of known mathematical constants.

PREREQUISITES
  • Understanding of infinite series and convergence
  • Familiarity with the properties of logarithms and trigonometric functions
  • Knowledge of the floor function and its implications in series
  • Basic proficiency in mathematical notation and manipulation
NEXT STEPS
  • Study the convergence of alternating series using the Alternating Series Test
  • Explore the derivation of $$\frac{\pi}{4}$$ through various series representations
  • Investigate the relationship between logarithmic functions and series expansions
  • Learn about the Euler-Maclaurin formula for approximating sums
USEFUL FOR

Mathematicians, students studying calculus or analysis, and anyone interested in advanced series evaluation techniques.

Saitama
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Evaluate the following:
$$\Large \sum_{k=1}^{\infty} (-1)^{\left\lfloor \frac{k+3}{2} \right\rfloor} \frac{1}{k}$$
 
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[sp]$$\begin{aligned}\sum_{k=1}^{\infty} (-1)^{\left\lfloor \frac{k+3}{2} \right\rfloor} \frac{1}{k} &= 1 + \frac12 - \frac13 - \frac14 + \frac15 + \frac16 - \frac17 -\frac18 + \ldots \\ &= \Bigl(1 - \frac13 + \frac15 - \frac17 + \ldots\Bigr) + \frac12\Bigl(1 - \frac12 + \frac13 - \frac14 + \ldots\Bigr) \\ &= \frac\pi4 + \frac12\ln2\end{aligned}$$[/sp]
 
Opalg said:
[sp]$$\begin{aligned}\sum_{k=1}^{\infty} (-1)^{\left\lfloor \frac{k+3}{2} \right\rfloor} \frac{1}{k} &= 1 + \frac12 - \frac13 - \frac14 + \frac15 + \frac16 - \frac17 -\frac18 + \ldots \\ &= \Bigl(1 - \frac13 + \frac15 - \frac17 + \ldots\Bigr) + \frac12\Bigl(1 - \frac12 + \frac13 - \frac14 + \ldots\Bigr) \\ &= \frac\pi4 + \frac12\ln2\end{aligned}$$[/sp]

Great as always! :)
 

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