Question

A series is defined by ∑n=1 to ∞ = 2/(2n−1)(2n+1). The partial sum of this series is [1]________.

Hence, this series is [2]________ . ​

Answer 1

The partial sum of the series ∑(n=1 to ∞) 2/[(2n−1)(2n+1)] is [1] 2. Hence, this series is [2] convergent.

To find the partial sum of the series
`\(\sum_(n=1)^(\infty) (2)/((2n-1)(2n+1))\)`, we can start by expressing the general term of the series:

`\[ a_n = (2)/((2n-1)(2n+1)) \]`

Now, let's find the partial sum up to the \(k\)-th term:

`\[ S_k = \sum_(n=1)^(k) a_n \]\[ S_k = \sum_(n=1)^(k) (2)/((2n-1)(2n+1)) \]`

Now, we can try to simplify this expression:

`\[ S_k = (2)/(1 \cdot 3) + (2)/(3 \cdot 5) + (2)/(5 \cdot 7) + \ldots + (2)/((2k-1)(2k+1)) \]`

To simplify this sum, notice that each term in the numerator and denominator has a common factor of 2. Let's factor that out:

`\[ S_k = 2 \left( (1)/(1 \cdot 3) + (1)/(3 \cdot 5) + (1)/(5 \cdot 7) + \ldots + (1)/((2k-1)(2k+1)) \right) \]`

Now, observe that each term in the parentheses follows a pattern:

`\[ (1)/((2n-1)(2n+1)) = (1)/(2n-1) - (1)/(2n+1) \]`

Therefore, the sum in the parentheses can be expressed as a telescoping series, where most terms cancel out:

`\[ (1)/(1 \cdot 3) + (1)/(3 \cdot 5) + (1)/(5 \cdot 7) + \ldots + (1)/((2k-1)(2k+1)) = \left(1 - (1)/(3)\right) + \left((1)/(3) - (1)/(5)\right) + \ldots + \left((1)/(2k-1) - (1)/(2k+1)\right) \]`

Notice that all the terms cancel out, except the first and the last:

`\[ S_k = 2 \left(1 - (1)/(2k+1)\right) \]`

Now, we can find the limit as \(k\) approaches infinity:

`\[ \lim_(k \to \infty) S_k = \lim_(k \to \infty) 2 \left(1 - (1)/(2k+1)\right) \]`

As k approaches infinity, the term
`\((1)/(2k+1)\)` approaches zero, so the limit simplifies to:

`\[ \lim_(k \to \infty) S_k = 2 \cdot 1 = 2 \]`

Therefore, the partial sum of the series is 2.

So, to answer your questions:

1. The partial sum of this series is
`\( \mathbf{2} \)`.

2. Hence, this series is
`\(\mathbf{\text{convergent}}\)`.