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.
n | nn+1 | an=(−1)nnn+1 |
1 | 12=0.5000 | −0.5000 |
2 | 23=0.6667 | 0.6667 |
3 | 34=0.7500 | −0.7500 |
4 | 45=0.8000 | 0.8000 |
5 | 56=0.8333 | −0.8333 |
6 | 67=0.8571 | 0.8571 |
7 | 78=0.8750 | −0.8750 |
8 | 89=0.8889 | 0.8889 |
9 | 910=0.9000 | −0.9000 |
10 | 1011=0.9091 | 0.9091 |
11 | 1112=0.9167 | −0.9167 |
12 | 1213=0.9231 | 0.9231 |
13 | 1314=0.9286 | −0.9286 |
14 | 1415=0.9333 | 0.9333 |
15 | 1516=0.9375 | −0.9375 |
16 | 1617=0.9412 | 0.9412 |
17 | 1718=0.9444 | −0.9444 |
18 | 1819=0.9474 | 0.9474 |
19 | 1920=0.9500 | −0.9500 |
20 | 2021=0.9524 | 0.9524 |
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.