How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #1/(1*2)+1/(2*3)+...+1/(n(n+1))+...#?

Answer 1

To find the nth partial sum of the series ( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \ldots + \frac{1}{n(n+1)} ), we first need to express each term in a more general form. Notice that ( \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} ).

So, the nth partial sum ( S_n ) can be written as:

[ S_n = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \ldots + \left( \frac{1}{n} - \frac{1}{n+1} \right) ]

Notice that most of the terms will cancel out, leaving only ( \frac{1}{1} - \frac{1}{n+1} ). Simplifying, we get:

[ S_n = 1 - \frac{1}{n+1} ]

To determine whether the series converges, we need to take the limit as ( n ) approaches infinity of the nth partial sum.

[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right) = 1 - 0 = 1 ]

Since the limit exists and is finite, the series converges.

To find the sum when it exists, we use the value of the limit:

[ \text{Sum} = 1 ]

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

#sum_(n=1)^oo 1/(n(n+1)) = 1#

Given the series:

#sum_(n=1)^oo 1/(n(n+1))#

we can determine that is convergent using the direct comparison test since:

#1/(n(n+1)) < 1/n^2#

and the series:

#sum_(n=1)^oo 1/n^2#
is convergent based on the #p#-series test.

Decompose now the general term using partial fractions:

#1/(n(n+1)) = 1/n-1/(n+1)#

and note that in the partial sum:

#s_n = (-1/(n+1) +1/n) + (-1/n +1/(n-1)) + ... + (-1/3+1/2) + (-1/2 +1)#

all terms cancel each other except the first and the last, so:

#s_n = 1-1/(n+1)#

and then:

#lim_(n->oo) s_n = 1#

is the sum of the series.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

The answer is #=1#

#sum_(k=1)^n1/(k(k+1))=sum_(k=1)^n1/k-sum_(k=1)^n1/(k+1)#
#=sum_(k=1)^n1/k-sum_(k=2)^(n+1)1/(k)#
#=1+cancel(sum_(k=2)^n1/k)-cancel(sum_(k=2)^n1/k)-1/(n+1)#
#=1-1/(n+1)#
#=n/(n+1)#
#lim_(n->+oo)n/(n+1)=lim_(n->+oo)1/(1+1/n)=1#

As

#lim_(n->+oo)1/n=0#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7