I have been stuck on this summation for almost a week now and I'd really appreciate some help. $$\sum_{r=1}^\infty \frac {r^3+(r^2+1)^2}{(r^4+r^2+1)(r^2+r)}$$
Infinite Summation $\sum_{r=1}^\infty \frac {r^3+(r^2+1)^2}{(r^4+r^2+1)(r^2+r)}$
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sequences-and-series
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2it would ne nice if you could add some own effort – 2017-01-14
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0adding $+r^2-r^2=0$ to the numerator seems to be a good idea – 2017-01-14
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6No idea whether it helps , but $$\frac{r^3+(r^2+1)^2}{(r^4+r^2+1)(r^2+r)}=\frac{1}{r}-\frac{1}{r+1}+\frac{1}{2(r^2-r+1)}-\frac{1}{2(r^2+r+1)}$$ – 2017-01-14
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1@Peter this essentialy solves the problem – 2017-01-14
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1@Peter. Please post this comment as a **good** answer. – 2017-01-14
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0@tired I just noticed $$(r+1)^2-(r+1)+1=r^2+r+1$$ In fact, we can take two telescope-sums – 2017-01-14
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1@Peter post this as an answer i will gladly upvote it – 2017-01-14
2 Answers
3
Hint : $$\frac{r^3+(r^2+1)^2}{(r^4+r^2+1)(r^2+r)}=\frac{1}{r}-\frac{1}{r+1}+\frac{1}{2(r^2-r+1)}-\frac{1}{2(r^2+r+1)}$$
and $$(r+1)^2-(r+1)+1=r^2+r+1$$
You get two telescope sums , the result is $\frac{3}{2}$
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HINT:
$$r^4+r^2+1=(r^2+1)^2-r^2=?$$
Use Partial Fraction Decomposition
$$\frac {r^3+(r^2+1)^2}{(r^4+r^2+1)(r^2+r)}=\dfrac ar+\dfrac b{r+1}+\dfrac{cr+d}{r^2-r+1}+\dfrac{er+f}{r^2+r+1}$$
Now if $f(r)=r^2-r+1, f(r+1)=?$