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The series in question is $$\sum_{n=1}^\infty\frac{n^2+1}{2n^2+5}$$

If $a_n$ is the $n$'th term of this sum, then $a_n \rightarrow \frac{1}{2}$ as $n\rightarrow \infty$. I believe this implies that the series doesn't converge - since this is effectively the negation of the statement "If a series converges, then $a_n$ tends to $0$ as $n$ tends to $\infty$". If I prove that the sequence $x_n=\frac{n^2 + 1}{2n^2 +5}$ converges to $\frac{1}{2}$, have I then proven that the series diverges? Or should I apply a convergence test and show that it doesn't pass one?

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    $\begingroup$ That will be sufficient $\endgroup$ – Paul Feb 27 at 11:50
  • $\begingroup$ Pertaining to the sequence or to the convergence test? $\endgroup$ – variations Feb 27 at 11:51
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    $\begingroup$ "If I prove that the sequence $x_n=\frac{n^2 + 1}{2n^2 +5}$ converges to $\frac{1}{2}$, have I then proven that the series diverges?" Yes you have. $\endgroup$ – Paul Feb 27 at 11:53
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The infinite series $$ \sum_{n=1}^\infty\frac{n^2+1}{2n^2+5}$$ diverges because the term $$\frac{n^2+1}{2n^2+5}\to 1/2 \not=0$$

Intuitively you are adding infinitely many numbers which are very close to $1/2$ and the result does not converge.

The so called divergence test indicates that if the general term of a series does not tend to zero, then the series diverges.

You do not have to prove anything else for the divergence of the above series.

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Just for your curiosity.

As already said, the problem is over as soon as you showed that the term tends to anything which is not zero.

If fact, we can even approximate the partial sums $$\frac{n^2+1}{2n^2+5}=\frac{1}{2}-\frac{3}{4 \left( n^2+\frac52\right)}$$ Now, use partial fraction decomposition $$\frac{1}{ n^2+\frac52}=\frac 1{(n+i\sqrt{\frac 52})(n-i\sqrt{\frac 52})}=\frac 1 {2a}\left(\frac 1{n-a} -\frac 1{n+a} \right)$$ $$\sum_{n=1}^p\frac 1{n+a} = H_{p+a}-H_a\qquad \text{and}\qquad \sum_{n=1}^p\frac 1{n-a} = H_{p-a}-H_{-a}$$ and using the asymptotics $$H_q=\gamma +\log \left({q}\right)+\frac{1}{2 q}-\frac{1}{12 q^2}+O\left(\frac{1}{q^3}\right)$$ After some minor simplifications of the complex numbers, you should end with $$S_p=\sum_{n=1}^p\frac{n^2+1}{2n^2+5}=\frac{p}{2}+\frac{3}{40} \left(2- \sqrt{10}\, \pi \,\coth \left(\sqrt{\frac{5}{2}} \pi \right)\right)+\frac{3}{4 p}-\frac{3}{8 p^2}+O\left(\frac{1}{p^3}\right)$$ Try it for $p=10$; the exact value is $$S_{10}=\frac{255436901298463}{57072223853645}\approx 4.47568 $$ while the truncated expansion given above leads to $4.47608$.

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