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$p$-Series and the Exact Value of $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$

The term $p$-series refers to series of the form $\displaystyle
\sum_{n=1}^{\infty} \frac{1}{n^p} =
\frac{1}{1^p} +\frac{1}{2^p}
+\frac{1}{3^p}+\cdots$. It can be easily shown in calculus, using the integral test, that the $p$-series diverges for $p\le1$ and converges for $p>1$. We have already seen that $p=1$ leads to a divergent series. Now we will finally investigate the Basel Problem, the $p$-series for $p = 2$.

  1. Since $\displaystyle\frac{1}{n^2}<\frac{2}{n(n+1)}$, show by using a telescoping sum how Jakob Bernoulli was able to prove that $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}<2$
  2. Euler showed that $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$.
    1. Find the exact sum of $\frac{1}{15}+\frac{1}{63}+\frac{1}{80}+\frac{1}{255}+\frac{1}{624}+\cdots$. ( The reciprocals of the numbers that are one less than perfect squares which simultaneously are other powers, i.e. $16= 4^2=2^4$ so $16-1=15$ is a denominator in the series.)
    2. Find the ratio of $\displaystyle
\sum_{n=1}^{\infty} \frac{1}{n^p}$ to $\displaystyle \sum_{n=1}^{\infty}
\frac{(-1)^{n+1}}{n^p}$ , where $p>1$.
    3. Find $\displaystyle
\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}$.
    4. Find $\displaystyle
\sum_{n=1}^{\infty} \frac{1}{n^4}$.



Zvezdelina Stankova-Frenkel 2002-03-24