How do you determine if #a_n=(1-1/8)+(1/8-1/27)+(1/27-1/64)+...+(1/n^3-1/(n+1)^3)+...# converge and find the sums when they exist?

Answer 1

The sum of the first #n# terms is given by:

# S_n = 1 - 1/(n+1)^3 #

The sum to infinity is given by:

# S = 1 #

Let us denote the #n^(th)# term of the sequence by:
# u_n = 1/n^3-1/(n+1)^3 #
Then we can denote the partial finite sum by #S_n# so that:
# S_n = sum_(r=1)^n \ u_r # # \ \ \ \ = (1-1/8)+(1/8-1/27)+... + (1/n^3-1/(n+1)^3)#
And then the infinite sum, by #S# where:
# S = sum_(r=1)^oo \ u_r #
First we can derive an expression for #S_n#, as we can write the sum of the first #n# terms as follows:
# S_n = u_1 + u_2 + ... u_n #
# \ \ \ \ \ = {1-1/8} + # # \ \ \ \ \ \ \ \ \ \ {1/8-1/27} + # # \ \ \ \ \ \ \ \ \ \ {1/27-1/64} + # # \ \ \ \ \ \ \ \ \ \ vdots # # \ \ \ \ \ \ \ \ \ \ {1/n^3-1/(n+1)^3 } #

This is a difference sum, and we can see that almost all terms will cancel with other terms:

# S_n = {1-cancelcolor(red)(1/8)} + # # \ \ \ \ \ \ \ \ \ \ {cancelcolor(red)(1/8)-cancelcolor(blue)(1/27)} + # # \ \ \ \ \ \ \ \ \ \ {cancelcolor(blue)(1/27)-cancel(1/64)} + # # \ \ \ \ \ \ \ \ \ \ vdots # # \ \ \ \ \ \ \ \ \ \ {(cancel(1/n^3)-1/(n+1)^3 } #

After which we are left with:

# S_n = 1 - 1/(n+1)^3 #
And for the sum, #S#, we have:
# S = lim_(n rarr oo) S_n # # \ \ = lim_(n rarr oo) 1 - 1/(n+1)^3 # # \ \ = 1-0 # # \ \ = 1 #
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Answer 2

To determine if the series ( a_n = (1 - \frac{1}{8}) + (\frac{1}{8} - \frac{1}{27}) + (\frac{1}{27} - \frac{1}{64}) + \ldots + (\frac{1}{n^3} - \frac{1}{(n+1)^3}) + \ldots ) converges and find the sums when they exist, we can analyze the series as a telescoping series.

  1. First, let's rewrite the terms of the series: [ a_n = 1 - \frac{1}{8} + \frac{1}{8} - \frac{1}{27} + \frac{1}{27} - \frac{1}{64} + \ldots + \frac{1}{n^3} - \frac{1}{(n+1)^3} + \ldots ]

  2. Notice that each term cancels out with the next term, except for the first and last terms: [ a_n = 1 - \frac{1}{(n+1)^3} ]

  3. Now, we can rewrite the series as: [ a_n = 1 - \frac{1}{(n+1)^3} ]

  4. To find out if the series converges, we need to take the limit of ( a_n ) as ( n ) approaches infinity: [ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(1 - \frac{1}{(n+1)^3}\right) ]

  5. Taking the limit, we get: [ \lim_{n \to \infty} \left(1 - \frac{1}{(n+1)^3}\right) = 1 - \lim_{n \to \infty} \frac{1}{(n+1)^3} = 1 - 0 = 1 ]

  6. Since the limit of ( a_n ) is a finite number (1), the series converges.

  7. To find the sum of the series, we sum up all the terms: [ S = \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \left(1 - \frac{1}{(n+1)^3}\right) ]

  8. This is a finite series, and we can find its sum using the formula for the sum of a finite geometric series: [ S = \frac{a_1}{1 - r} ] where ( a_1 ) is the first term and ( r ) is the common ratio.

  9. In this series, ( a_1 = 1 ) and ( r = 0 ) (since each term cancels out with the next term).

  10. Substituting the values into the formula, we get: [ S = \frac{1}{1 - 0} = \frac{1}{1} = 1 ]

Thus, the sum of the series is 1.

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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.

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