I have an infinite arithmetic series whose terms are expressed as $$\sum _{k = 1}^{\infty }\frac{2k^2+3}{k} $$ and the sum of it is equal to $$(n + 1)n + 3H_n$$ I can't wrap my head around how they got to this answer.
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In the title you have the sum going to infinity. In the result of the summation you have $n$'s appearing. In any event, you should try to make an attempt to format this with MathJax if possible. – JMoravitz May 22 '20 at 13:42
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On to the math involved here... recall that $\sum\limits_{i=1}^n (a_i+b_i) = \sum\limits_{i=1}^n a_i + \sum\limits_{i=1}^n b_i$, so you can split your sum into two. The first sum $\sum\limits_{i=1}^n \dfrac{2i^2}{i}$ simplifies to $2\sum\limits_{i=1}^n i$ which you should recognize is twice the $n$'th triangular number and so $(n+1)n$. The second part $\sum\limits_{i=1}^n \dfrac{3}{i}=3\sum\limits_{i=1}^n \dfrac{1}{i}$ is $3H_n$ by definition of $H_n$. – JMoravitz May 22 '20 at 13:45
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@JMoravitz Thank you, this largely answers my question. I still don't understand, however, why 2 sum(i = 1 to n) i is equal to (n+1)n. Is there a general formula for solving these kind of problems? – Joseph Rikman May 22 '20 at 13:47
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Proof $1+2+3+4+\dots+n = \frac{n\times (n+1)}{2}$ – JMoravitz May 22 '20 at 13:49
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@JMoravitz Is it usual procedure to provide a full answer to questions like this, or just a hint towards the first step? Because my approach is to offer OP the first step, or a hint towards the insight needed, or whatever. (Plus I'm too lazy to plod through all the details, that's OP's job.) – Prime Mover May 22 '20 at 13:57
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Given the confusion regarding what the question is actually saying (is it going to infinity or to $n$?) and the inclarity based on the formatting of the expression, all I will do is offer a clue.
Try separating the expression in brackets into two separate summations: one of $\dfrac {2 i^2} i = 2i$ and one of $\dfrac 3 i$.
You will find it's a lot simpler than it looked at first.
Prime Mover
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