Limits of Sums (Problem 1)

Algebra Level 2

Determine the value of $\lim_{n\to\infty} \displaystyle \sum_{i=1}^n \frac{1}{i(i+1)}.$

The answer is 1.

2 solutions

JohnDonnie Celestre
Apr 25, 2014

Decompose $\frac{1}{i(i+1)}$ into partial fractions, that is: $\frac{1}{i} - \frac{1}{i+1}$ . $\lim_{n\to\infty}\sum_{i=1}^n (\frac{1}{i} - \frac{1}{i+1})$ Now, substitute values for $i$ from $1$ to $n$ . $\lim_{n\to\infty}((\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + ... + (\frac{1}{n} - \frac{1}{n+1}))$ $\lim_{n\to\infty}(\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{n} - \frac{1}{n+1})$ $\lim_{n\to\infty}(\frac{1}{1} - \frac{1}{n+1})$ $\lim_{n\to\infty}(1 - \frac{1}{n+1})$ $\lim_{n\to\infty}1 - \lim_{n\to\infty}\frac{1}{n+1}$ $1-0$ $\boxed{1}$

Why can you separate the fractions like that vice (1/i)(1/(i+1)). Where does the subtraction come from?

Charity Reed - 4 years, 7 months ago

We assume that $\frac{1}{i}\cdot\frac{1}{i+1}$ is equal to some expression of the form $\frac{A}{i} + \frac{B}{i+1}$ , where $A$ and $B$ are both constants. Multiplying both sides of this equation by the factor $i(i+1)$ , we obtain $1 = A(i+1) + B(i)$ . We substitute simple values of i to see if we can obtain some useful equations: by substituting $i = 0$ , we obtain $A = 1$ ; and by substituting $i = -1$ , we obtain $B = -1$ . Thus, we conclude that $\frac{1}{i}\cdot \frac{1}{i+1}$ is equal to $\frac{1}{i} - \frac{1}{i+1}$ . As a check, you can multiply both sides of the equation $\frac{1}{i}\cdot \frac{1}{i+1} = \frac{1}{i} - \frac{1}{i+1}$ by $i(i+1)$ to obtain the tautology $1 = (1+i) - i = 1$ .

Ed Karabinus - 4 years, 6 months ago

(1/i)x(1/(i+1))=/=(A/i)+(B/(i+1)) as 1=1 but A does not necessarily need to equal B, thus the answer is invalidated.

Gerasimos Konidaris - 4 years, 4 months ago

I think this is wrong. 1/i is an harmonic serie, so the summation of 1/i = infinity.

Jean-Bruno Boulianne - 4 years, 5 months ago

Why does 1/i not go to infinite?

Vini Mehta - 4 years, 7 months ago

as i goes to infinite, 1/i goes to zero.

Mark Shigley - 4 years, 7 months ago

Partial Fraction Decomposition is how you split up the initial expression.

Aviv Brafman - 4 years, 6 months ago

Karen Vardanyan
Jun 9, 2014

Another solution appears when you remember that $\sum _{ i=1 }^{ n }{ \frac { 1 }{ n(n+1) } } =1-\frac { 1 }{ n+1 }$ .

could please explain how? (i know sorry for the dumb question)

Erfan Huq - 4 years, 4 months ago

Limits of Sums (Problem 1)

The answer is 1.

2 solutions

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