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- Thread starter jeca86
- Start date
Hello, jeca86!
How about a Comparison?
Hence: \(\displaystyle \L\,\frac{n^2\,-\,1}{3n^4\,+\,1}\:<\:\frac{n^2}{3n^4}\:=\:\frac{1}{3n^2}\)
Then: \(\displaystyle \L\:\sum^{\infty}_{n=1}\frac{n^2\,-\,1}{3n^4\,+\,1}\;<\;\frac{1}{3}\sum^{\infty}_{n=1}\frac{1}{n^2}\)
And the latter is a convegent p-series.
How about a Comparison?
Note that: \(\displaystyle \,n^2\,-\,1\:<\:n^2\) and \(\displaystyle 3n^4\,+\,1\:>\:3n^4\)convergent or divergent?\(\displaystyle \L\:\sum^{\infty}_{n=1}\frac{n^2\,-\,1}{3n^4\,+\,1}\)
Hence: \(\displaystyle \L\,\frac{n^2\,-\,1}{3n^4\,+\,1}\:<\:\frac{n^2}{3n^4}\:=\:\frac{1}{3n^2}\)
Then: \(\displaystyle \L\:\sum^{\infty}_{n=1}\frac{n^2\,-\,1}{3n^4\,+\,1}\;<\;\frac{1}{3}\sum^{\infty}_{n=1}\frac{1}{n^2}\)
And the latter is a convegent p-series.
If I may add my 2 cents. For the curious. The convergent p-series that Soroban pointed out is one of the best known convergent series.
\(\displaystyle \L\\\sum_{n=1}^{\infty}\frac{1}{n^{2}}\)
Euler proved it converges to \(\displaystyle \frac{{\pi}^{2}}{6}\)
Using the sine series, which he regarded as an "infinite polynomial", his method was ingenious.
\(\displaystyle \L\\\sum_{n=1}^{\infty}\frac{1}{n^{2}}\)
Euler proved it converges to \(\displaystyle \frac{{\pi}^{2}}{6}\)
Using the sine series, which he regarded as an "infinite polynomial", his method was ingenious.