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$\begingroup$

I tried simplifying it by first resolving it into partial fractions but to no use. Also I know that $\sum (1/n)$ is harmonic number, but I don't know how to use it. Please help me.

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    $\sum_{n=1}^{2017}\frac{1}{n^{2}+2n}=\sum_{n=1}^{2017}\frac{1}{n(n+2)}=\frac{1}{2}\sum_{n=1}^{2017}\left(\frac{1}{n}-\frac{1}{n+2}\right)$. Note that this is telescoping. – 2017-01-04
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    Well thanks for that last expression.... I'll try singing with it.... Else please help me – 2017-01-04
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    This may help:$$\sum_{n=1}^{2017}\frac{1}{n+2}=\sum_{n=3}^{2019}\frac{1}{n}$$ What does the expression "singing with it" mean? I've never heard that before. – 2017-01-04
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    "singing with it", sorry for this...I wanted to type "something with it", but since I use glide typing I might have pressed the wrong buttons. – 2017-01-04
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    @LokeshSangewar LOL singing with it... – 2017-01-04
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    But then again sigma 1/n n=3 to 2019.....How to go ahead with this....Coz I want the answer as a single rational – 2017-01-04
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    $$\sum_{n=1}^{2017}\left(\frac{1}{n}-\frac{1}{n+2}\right)=\sum_{n=1}^{2017}\frac{1}{n}-\sum_{n=3}^{2019}\frac{1}{n}$$ Note that many terms cancel. – 2017-01-04
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    Okkkk....I got it. Thanks a lot!!! – 2017-01-04
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    @LokeshSangewar No problem. – 2017-01-04
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    But dude....(1/n)-(1/n+2) is not equal to 1/n(n+2).....So how to do it??????? – 2017-01-04
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    @Servaes nope......I don't think so – 2017-01-04
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    @Servaes Did I make an error? I'd be happy to delete my comments if this has happened. – 2017-01-04
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    @user71352 well.... Unfortunately yes – 2017-01-04
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    @LokeshSangewar Oh shoot! Can you point me to the error? – 2017-01-04
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    U wrote 1/n(n+2) as 1/n - 1/n+2.....That's wrong..... – 2017-01-04
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    @LokeshSangewar I gather, perhaps incorrectly, you are referring to my first comment. I believe there is a factor of $\frac{1}{2}$ on the outside of the sum and $$\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)=\frac{1}{2}\cdot\frac{(n+2)-n}{n(n+2)}=\frac{1}{n(n+2)}$$ – 2017-01-04
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    @user71352 You didn't. I was snidely commenting on the fact that your first three comments very explicitly give a step by step answer, yet the OP asks a question with an extremely obvious answer that above all you already gave. The OP's homework has now been done without any understanding from his/her side, which bothers me. But expressing this in such a way might not be helpful, so I'll delete my comment. – 2017-01-04

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$\displaystyle\sum_{k=1}^{2017}\dfrac 1{n^2+2n}=\frac{1}{2}\sum_{k=1}^{2017}(\frac {1}{n}-\frac{1}{n+2})$

Writing it out:

$\frac{1}{2}[1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ -(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019})]$

$=\frac{1}{2}(1+\frac{1}{2}-\frac{1}{2018}-\frac{1}{2019})$

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    I could be mistaken but I think there should be negatives on $\frac{1}{2018}$ and $\frac{1}{2019}$. – 2017-01-04
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    Yh that's correct.... They're negative – 2017-01-04
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    Well I have posted another question....It has not been answered yet....Well if anyone can help me with that – 2017-01-04
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    @user71352 Rectified. Thanks. – 2017-01-04
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    @WiCK3DPOiSON No problem. (+1) – 2017-01-04