If $\large\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}n=a$ then what is the value of $\large\sum_{n=1}^{\infty}\frac1n\cos\frac{n\pi}2$ ?
$1)\frac{-a_1}2\qquad\qquad2)-\frac a2\qquad\qquad3)\frac{a-a}2\qquad\qquad4)\frac a2$
To solve this problem I just evaluated each sum:
$$a=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}n=\frac11-\frac12+\frac13-\frac14+\cdots$$ $$\sum_{n=1}^{\infty}\frac1n\cos\frac{n\pi}2=-\frac12+\frac14-\frac16+\cdots$$ Then noticed that if I multiply the terms of the first sums by $-\frac12$ It gives me the second sum. so the answer is$-\frac a2$.
But is it possible to solve this question without writing and adding the terms of each sum and comparing the terms?