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$$\displaystyle\sum_{k=1}^n \frac1{k(k+1)}$$

Method of Exhuastion for the area of a parabolic segment | Physics Forums

Hello everyone, I'm using the book Apostol- Calculus Vol. 1 for self-studying to get a better understanding of proof based Calculus. They said this book was good for self studying, but I am already stuck in the first chapter. I'm trying to understand how he got the identity: 12+22+...+n(2)=...
$$\sum_{k=1}^{n} \frac{1}{k(k+1)}$$

Explaining the pattern

https://math.stackexchange.com/questions/3686497/explaining-the-pattern#comment-7575876
[Evaluating \sum_{k=1}^{n} \frac{1}{k(k+1)} ](https://math.stackexchange.com/questions/3256519/evaluating-sum-k-1n-frac...
$$\displaystyle\sum_{k=1}^{n} \frac 1 {k(k + 1)}$$

Compute $\sum_{k=1}^{n} \frac 1 {k(k + 1)}$

https://math.stackexchange.com/questions/293244/compute-%24%5Csum_%7Bk%3D1%7D%5E%7Bn%7D-%5Cfrac-1-%7Bk%28k-%2B-1%29%7D-%24
More specifically, I'm supposed to compute \displaystyle\sum_{k=1}^{n} \frac 1 {k(k + 1)} by using the equality \frac 1 {k(k + 1)} = \frac 1 k - \frac 1 {k + 1} and the problem before which just says that, \displaystyle\sum_{j=1}^{n} a_j - a_{j - 1} = a_n - a_0 . I can add up the sum for any n but I'm not sure what they mean by "compute". Thanks!...
$$\sum_{k=1}^n \frac{1}{k(k+1)}$$

Proof Question: Using Mathematical Induction | Physics Forums

Homework Statement Prove that, for all integers n =>1 \frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}... + \frac{1}{n+1} = 1-\frac{1}{n+1} Homework Equations I am a little stuck on this question. :| The Attempt at a Solution...
5\sum_{k=1}^n\frac{1}{k(k+1)}

How do I write (5/(1*2))+(5/(2*3))+(5/(3*4))+...+(5/n(n+1))+...in summation notation, and how can I tell if the series converges? | Socratic

https://socratic.org/questions/how-do-i-write-5-1-2-5-2-3-5-3-4-5-n-n-1-in-summation-notation-and-how-can-i-tel
5\sum_{k=1}^n\frac{1}{k(k+1)} The given series: (5/{1\cdot 2})+(5/{2\cdot 3})+(5/{3\cdot 4})+\ldots+(5/{n(n+1)}) =\sum_{k=1}^n\frac{5}{k(k+1)} =5\sum_{k=1}^n\frac{1}{k(k+1)} =5\sum_{k=1}^n(1/k-\frac{1}{k+1}) =5((1-1/2)+(1/2-1/3)+(1/3-1/4)+\ldots+(1/n-1/{n+1})) =5(1-1/2+1/2-1/3+1/3-1/4+\ldots+1/n-1/{n+1}) =5(1-1/{n+1}) \therefore \lim_{n\to \infty}\sum_{k=1}^n\frac{5}{k(k+1)} =\lim_{n\to \infty}5(1-1/{n+1}) =5(1-0) =5 Hence, the given se...
$$S_1 = \sum_{k=1}^n {1 \over k(k+1)},$$

Find the sum to n terms of the series $\frac{1} {1\cdot2\cdot3\cdot4} + \frac{1} {2\cdot3\cdot4\cdot5} + \frac{1} {3\cdot4\cdot5\cdot6}\ldots$

https://math.stackexchange.com/questions/5558/find-the-sum-to-n-terms-of-the-series-%24%5Cfrac%7B1%7D-%7B1%5Ccdot2%5Ccdot3%5Ccdot4%7D-%2B-%5Cfrac%7B1%7D-%7B2%5Ccdot3%5Ccdot4%5Ccdot5%7D-%2B-%5Cfrac%7B1%7D-%7B3%5Ccdot4%5Ccdot5%5Ccdot6%7D%5Cldots-%24#5572
This is similar to what has been said by Branimir, but shows how we can extend the result to \sum_{k=1}^n {1 \over k(k+1) \cdots (k+m)}, \qquad m \in \mathbb{N}. We can build up the result from the identities {1 \over k(k+1)} = {1 \over k} - { 1 \over k+1}, \qquad (1) {1 \over k(k+1)(k+2)} = {1 \over 2} \left( {1 \over k(k+1)} - { 1 \over (k+1)(k+2)} \right), {1 \over k(k+1)(k+2)(k+3)} = {1 \over 3} \left( {1 \over k(k+1)...
$$\displaystyle \ \sum_{k=1}^n \frac{1}{k(k+1)} \,.\$$

Express this sum as a fraction of whole numbers | Physics Forums

Homework Statement Express the sum as a fraction of whole numbers in lowest terms: ##\frac{1}{1⋅2}+\frac{1}{2⋅3}+\frac{1}{3⋅4}+...+\frac{1}{n(n+1)}## Homework Equations The Attempt at a Solution Please see attached image for my work. The reason I am posting the image rather than typing this...
$$\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}$$

Prove that $\sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$

https://math.stackexchange.com/questions/1837431/prove-that-%24%5Csum_%7Bx%3D1%7D%5E%7Bn%7D-%5Cfrac%7B1%7D%7Bx-%28x%2B1%29%28x%2B2%29%7D-%3D-%5Cfrac%7B1%7D%7B4%7D---%5Cfrac%7B1%7D%7B2-%28n%2B1%29-%28n%2B2%29%7D%24#comment-5287775
$$\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)}$$

$\sum _{n=1}^{\infty } \frac{1}{n (n+1) (n+2)}$ Understand the representation

https://math.stackexchange.com/questions/1108626/%24%5Csum-_%7Bn%3D1%7D%5E%7B%5Cinfty-%7D-%5Cfrac%7B1%7D%7Bn-%28n%2B1%29-%28n%2B2%29%7D%24-understand-the-representation#1108660
You should write: (define s_n as the sequence of partial sums of the series) s_n=\sum _{k=1}^{n} \frac{1}{k (k+1) (k+2)} =\sum _{k=1}^{n} {\frac{1}{2} \left(-\frac{2}{k+1}+\frac{1}{k+2}+\frac{1}{k}\right)} =\frac{1}{2}\left(-2\sum_{k=1}^{n}\frac{1}{k+1}+\sum_{k=1}^{n}\frac{1}{k+2}+\sum_{k=1}^{n}\frac{1}{k}\right) because the sums are finite. Observe then that the three sums overlap on most terms. =\frac{1}{2}\left(-2\sum_{k=2}^...
$$\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}$$

series: $\frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots$

https://math.stackexchange.com/questions/2110348/series%3A-%24%5Cfrac%7B1%7D%7B2%5Ccdot-3%5Ccdot-4%7D%2B%5Cfrac%7B1%7D%7B4%5Ccdot-5%5Ccdot-6%7D%2B%5Cfrac%7B1%7D%7B6%5Ccdot-7%5Ccdot-8%7D%2B%5Ccdots%24
We have the infinite series: \frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots This is not my series: \frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{4\cdot 5\cdot 6}+\cdots so I cannot use \sum_{k=1}^n \frac{1}{k(k+1)(k+2)} My attempt: I know that this type of series solved by making it telescoping series but here, I am unable find general term of the ser...
$$a_n=\sum_{k=1}^n \frac {1}{k(k+1)^2}$$

Does $a_n = \frac {1}{1*2^2}+\frac {1}{2*3^2}+...+\frac{1}{n*(n+1)^2}$ converge?

https://math.stackexchange.com/questions/3417009/does-%24a_n-%3D-%5Cfrac-%7B1%7D%7B1*2%5E2%7D%2B%5Cfrac-%7B1%7D%7B2*3%5E2%7D%2B...%2B%5Cfrac%7B1%7D%7Bn*%28n%2B1%29%5E2%7D%24-converge%3F#comment-7027492
Maybe you mean \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac {1}{n(n+1)^2} ? or a_n=\sum_{k=1}^n \frac {1}{k(k+1)^2}...
$$\sum_{k=1}^n\frac{1}{k(k+1)...(k+m)}$$

A proof for the identity $\sum_{n=r}^{\infty} {n \choose r}^{-1} = \frac{r}{r-1}$

https://math.stackexchange.com/questions/2124113/a-proof-for-the-identity-%24%5Csum_%7Bn%3Dr%7D%5E%7B%5Cinfty%7D-%7Bn-%5Cchoose-r%7D%5E%7B-1%7D-%3D-%5Cfrac%7Br%7D%7Br-1%7D%24#2124251
This is basically a generalization of the approach from this answer: begin{align*} \sum_{n=r}^\infty \frac1{\binom nr} &= r! \sum_{n=r}^\infty \frac1{n(n-1)\cdots(n-r+1)}\\ &= \frac{r!}{r-1} \sum_{n=r}^\infty \frac{n-(n-r+1)}{n(n-1)\cdots(n-r+1)}\\ &= \frac{r!}{r-1} \sum_{n=r}^\infty \left(\frac1{(n-1)\cdots(n-r+1)} - \frac1{n(n-1)\dots(n-r+2)}\right)\\ &\overset{(1)}= \frac{r!}{r-1} \cdot \frac1{(r-1)!} = \\ &= \fr...
$$\sum_{k=1}^n\frac{1}{k(k+1)...(k+m)}$$

General formula for this sum $\sum_{k=1}^n\frac{1}{k(k+1)...(k+m)}$

https://math.stackexchange.com/questions/2003497/general-formula-for-this-sum-%24%5Csum_%7Bk%3D1%7D%5En%5Cfrac%7B1%7D%7Bk%28k%2B1%29...%28k%2Bm%29%7D%24
Is there a general formula for \sum_{k=1}^n\frac{1}{k(k+1)...(k+m)} ? I know that the limit is \frac{1}{mm!} but is there a combinatorial expression for this?...
$$S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}$$

Calculate $S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}$.

https://math.stackexchange.com/questions/1109806/calculate-%24s-%3D%5Csum_%7Bk%3D1%7D%5En%5Cfrac-%7B1%7D%7Bk%28k%2B1%29%28k%2B2%29%7D%24.
Calculate S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)} . I know I posted this question already but I want a more detailed answer. For example, how you got from one step to another using the partial fraction formula. Thanks!...
$$S =\displaystyle\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}$$

Calculate $S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}.$

https://math.stackexchange.com/questions/1109391/calculate-%24-s-%3D%5Csum_%7Bk%3D1%7D%5En%5Cfrac-%7B1%7D%7Bk%28k%2B1%29%28k%2B2%29%7D.-%24
Calculate S =\displaystyle\sum_{k=1}^n\frac {1}{k(k+1)(k+2)} . This sequence is neither arithmetic nor geometric. How can you solve this. Thanks!...
$$\sum\limits_{k=1}^{n}\frac{1}{k(k+1)}=\frac{n}{n+1}$$

Mathematical Induction, Want to check I'm getting this right

https://math.stackexchange.com/questions/2055816/mathematical-induction%2C-want-to-check-i%27m-getting-this-right#2055839
First, show that this is true for n=1 : \sum\limits_{k=1}^{1}\frac{1}{k(k+1)}=\frac{1}{1+1} Second, assume that this is true for n : \sum\limits_{k=1}^{n}\frac{1}{k(k+1)}=\frac{n}{n+1} Third, prove that this is true for n+1 : \sum\limits_{k=1}^{n+1}\frac{1}{k(k+1)}= \color\red{\sum\limits_{k=1}^{n}\frac{1}{k(k+1)}}+\frac{1}{(n+1)(n+1+1)}= \color\red{\frac{n}{n+1}}+\frac{1}{(n+1)(n+1+1)}= \frac{n+1}{n+1+1} Please note tha...
$$\sum_{k=1}^n \dfrac1{k(k+1)} =\dfrac{n}{n+1}$$

How to prove using induction that $\sum_{i = 1}^n \frac{1}{\sum_{n = 0}^i n} = \frac{2n}{n+1}$

https://math.stackexchange.com/questions/2022169/how-to-prove-using-induction-that-%24-%5Csum_%7Bi-%3D-1%7D%5En-%5Cfrac%7B1%7D%7B%5Csum_%7Bn-%3D-0%7D%5Ei-n%7D-%3D-%5Cfrac%7B2n%7D%7Bn%2B1%7D%24#2022187
To do this by induction, I start at the OP's simplification \sum_{k=1}^n \dfrac1{k(k+1)} =\dfrac{n}{n+1} . This is true for n=1 , where it says \frac12 = \frac12 . If it is true for n , then begin{array}\\ \sum_{k=1}^{n+1} \dfrac1{k(k+1)} &=\sum_{k=1}^{n} \dfrac1{k(k+1)}+\dfrac1{(n+1)(n+2)}\\ &=\dfrac{n}{n+1}+\dfrac1{(n+1)(n+2)}\\ &=\dfrac{n(n+2)+1}{(n+1)(n+2)}\\ &=\dfrac{n^2+2n+1}{(n+1)(n+2)}\\ &=\dfrac{(n+1)...
$$\displaystyle \sum_{k=1}^n \frac{1}{k(k+1)} = \frac{n}{n+1}$$

How to proof that $\forall n \ge 1: \sum_{i=1}^n (i(i+1))^{-1} = n(n+1)^{-1}$ using mathematical induction

https://math.stackexchange.com/questions/2464297/how-to-proof-that-%24%5Cforall-n-%5Cge-1%3A-%5Csum_%7Bi%3D1%7D%5En-%28i%28i%2B1%29%29%5E%7B-1%7D-%3D-n%28n%2B1%29%5E%7B-1%7D%24-using-mathematical-induction#2464305
\sum_{k=1}^n \frac{1}{k(k+1)} = \frac{n}{n+1} Suppose that the above is true for some n , then for n+1 we have: \sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{1}{(n+1)(n+2)} + \sum_{k=1}^{n} \frac{1}{k(k+1)} = \frac{1}{(n+1)(n+2)} +\frac{n}{n+1} Simplifying we get: \frac{1}{(n+1)(n+2)} +\frac{n}{n+1} = \frac{n^2 + 2n + 1}{(n+1)(n+2)} = \frac{(n+1)^2}{(n+1)(n+2)} = \frac{n+1}{n+2} So if \displaystyle \sum_{k=1}^n \frac{1}{k(k+...
$$\sum_{k=1}^{n}\frac{1}{k(k+1)}=\frac{n}{n+1}$$

How Are the Solutions for Finite Sums of Natural Numbers Derived?

https://math.stackexchange.com/questions/2664218/how-are-the-solutions-for-finite-sums-of-natural-numbers-derived%3F#2664224
Note that \sum_{k=1}^{n}k=\frac{n(n+1)}{2} is a classical result which can be easily proved by the following trick and also \sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}=\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4} can be derived by a similar trick in 3D Note that \sum_{k=1}^{n}k(k+1)=\frac{n(n+1)(n+2)}{3}=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} is simply \sum_{k=1}^{n}k(k+1)=\sum_{k=1}^{n}k^2+\sum_{k=1}^{n}k and \sum_{k=1}^...
$$\sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=2}^{n+1}\frac{1}{k(k-1)}$$