Question

Determine whether the series is convergent or divergent by expressing sₙ as a telescoping sum

∑ₙ=1[infinity]21/n(n+3)

Answer 1

To determine if the series ∑ₙ=1^∞ 21/n(n+3) converges or diverges, we express it as a telescoping sum. By rewriting the terms and observing their cancellations, we find that the series converges; Therefore the given series is Convergent.

Step-by-step explanation:

To determine if the series ∑ₙ=1^∞ 21/n(n+3) is convergent or divergent, we can express it as a telescoping sum.

Write out the terms of the series for the first few values of n:

For n = 1: 21/1(1+3) = 21/4

For n = 2: 21/2(2+3) = 21/10

For n = 3: 21/3(3+3) = 21/18

For n = 4: 21/4(4+3) = 21/28

...

We can see that the terms of the series have the form 21/n(n+3). We want to find a telescoping sum given by Sₙ = a₁ - a₂ + a₂ - a₃ + a₃ - a₄ + ...

To rewrite the terms in this form, we split 21/n(n+3) into partial fractions:

21/n(n+3) = A/n + B/(n+3)

To solve for A and B, we find the common denominator:

A(n+3) + Bn = 21

Setting n = -3 gives:

A(0) + B(-3) = 21

-3B = 21

B = -7

Setting n = 0 gives:

A(3) + B(0) = 21

3A = 21

A = 7

Therefore, we can express 21/n(n+3) as:

21/n(n+3) = 7/n - 7/(n+3)

Now, Rewrite the series using these expressions:

∑ₙ=1^∞ 21/n(n+3) = ∑ₙ=1^∞ 7/n - 7/(n+3)

We can observe the telescoping nature of the series because each consecutive pair of terms cancels each other out. Simplify the series:

(7/1 - 7/4) + (7/2 - 7/5) + (7/3 - 7/6) + (7/4 - 7/7) + ...

Each term cancels with the following term, except for the leading term of the first term (7/1) and the trailing term of the last term (-7/(n+3)). Si, the sum will be:

Sₙ = 7/1 - 7/(n+3)

Now, Considering the behavior of Sₙ as n approaches infinity:

limₙ→∞ Sₙ = limₙ→∞ (7/1 - 7/(n+3))

= 7/1 - 7/∞

= 7

Since the limit of the partial sums, Sₙ, exists and is finite as n approaches infinity, the series ∑ₙ=1^∞ 21/n(n+3) converges.

Therefore, the given series is convergent.

Answer 2

By decomposing the term into partial fractions and writing out the partial sum, it is shown that the series ∑ₙ=1[∞]21/n(n+3) diverges as the positive terms approach zero and the negative terms form a divergent harmonic series.

Step-by-step explanation:

To determine whether the series ∑ₙ=1[∞]21/n(n+3) is convergent or divergent by expressing sₙ as a telescoping sum, we first need to decompose the general term into partial fractions. We want to write 21/(n(n+3)) as A/n + B/(n+3) for some constants A and B. By finding a common denominator and equating coefficients, we can solve for A and B.

Let's do the decomposition:

21/(n(n+3)) = A/n + B/(n+3)

Multiplying both sides by n(n+3) to clear out the denominators, we then get:

21 = A(n+3) + Bn

Setting n = 0, we find B = 7. Setting n = -3, we find A = -7. So the decomposition is:

21/(n(n+3)) = -7/n + 7/(n+3)

Now, we write the partial sum sₙ for n terms:

sₙ = ∑ₙ=1^n (-7/n + 7/(n+3))

As we sum up these terms, most will cancel out in a telescoping fashion:

7/1 + 7/4

-7/2 + 7/5

-7/3 + 7/6

...

-7/(n-3) + 7/n

-7/(n-2) + 7/(n+1)

-7/(n-1) + 7/(n+2)

-7/n + 7/(n+3)

It's clear that every term except for the first few and the last few will cancel out. The remaining terms will be -7/1 - 7/2 - 7/3 + 7/(n+1) + 7/(n+2) + 7/(n+3) as n approaches infinity. Now we can see that as n approaches infinity, the positive terms will approach zero and the sum of the negative terms diverges (since the sum 1/n is a known divergent harmonic series). Therefore, the whole series diverges.