Whether the sequence is convergent or divergent; guess the value of the limit if the sequence converges with valid proof.

The sequence is divergent.

Given:

The sequence is an=(−1)nnn+1.

Definition used:

If an is a sequence and limn→∞an exists, then the sequence an is said to be convergent otherwise it is divergent.

Calculation:

Obtain the first 20 terms of the sequence.

Plot the points (n,an), for n=1,2,...20 on the graph as shown below in Figure 1.

From the graph, it is observed that the sequence is divergent because the sequence oscillates between −1 and 1.

To prove:

The sequence an=(−1)nnn+1 is divergent.

Proof:

Obtain the limit of the sequence (the value of the term an as n tends to infinity).

That is, compute limn→∞an=limn→∞((−1)nnn+1).

Use the subsequence n=2m in equation (1).

limn→∞((−1)nnn+1)=limm→∞((−1)2m⋅2m2m+1)=limm→∞(((−1)2)m⋅2m2m+1)=limm→∞(1m⋅2m2m+1)=limm→∞(2m2m+1)

Divide the numerator and denominator by the highest power.

limm→∞((−1)2m⋅2m2m+1)=limm→∞(2mm2m+1m)                                =limm→∞(22+1m)                                =limm→∞(2)limm→∞(2+1m)=limm→∞(2)limm→∞(2)+limm→∞(1m)

Apply infinity property limn→∞(kna)=0 and simplify the terms.

limm→∞((−1)2m⋅2m2m+1)=22+1∞=22+0=22=1

Therefore, limn→∞((−1)nnn+1)=1 when n=2m.

Use the subsequence n=(2m+12)π in equation (1).

limn→∞((−1)nnn+1)=limm→∞((−1)2m+1⋅2m+12m+1+1)=limm→∞(−((−1)2)m⋅2m2m+2)=limm→∞(−1m⋅2m2m+2)=limm→∞(−2m2m+2)

Divide the numerator and denominator by the highest power.

limm→∞((−1)2m⋅2m2m+1)=limm→∞(−2mm2m+2m)                                =limm→∞(−22+2m)                                =limm→∞(−2)limm→∞(2+2m)=−2limm→∞(2)+limm→∞(2m)

Apply infinity property limn→∞(kna)=0 and simplify the terms.

limm→∞((−1)2m⋅2m2m+1)=−22+1∞=−22+0=−22=−1

Therefore, limn→∞((−1)nnn+1)=−1 when n=2m+1.

Thus, the term an oscillates between −1 and 1.

Since limn→∞an does not exist and by the definition of the sequence, it can be concluded that the sequence is divergent.

Therefore, the sequence is divergent.