To determine
To calculate: The first 10 partial sums terms of the series and plot the sequence of terms and the sequence of partial sums on the graph to obtain the sum if the series convergent.
Answer
The first 10 partial sums of the series are 1.00000, 1.33333, 1.50000, 1.60000, 1.66667, 1.71429, 1.75000, 1.77778, 1.80000 and 1.81818.
The graph of the sequence of terms and the sequence of partial sums is shown below in Figure 1.
The series is convergent to the sum 2.
Explanation
Given:
The series is
∑n=2∞2n2−n
.
Here, the sequence of the term is
an=2n2−n
.
Result used:
The sum of the geometric series is
∑n=1∞arn−1=a1−r if r<1
.
Calculation:
The nth term of the partial sum is
sn=∑i=2nai
.
Obtain the first 10 terms of the sequence and partial sums.
n 
an=2n2−n

sn=∑i=2nai

2 
a2=1

s2=1

3 
a3=0.333333

s3=1.333333

4 
a4=0.166667

s4=1.500000

5 
a5=0.100000

s5=1.600000

6 
a6=0.066667

s6=1.666667

7 
a7=0.047619

s7=1.714286

8 
a8=0.035714

s8=1.750000

9 
a9=0.027778

s9=1.777778

10 
a10=0.022222

s10=1.800000

11 
a11=0.018182

s11=1.818182

Therefore, the first 10 terms of the sequence are 1.00000, 0.33333, 0.166667, 0.10000,
0.06667, 0.04762, 0.03571, 0.02778, 0.02222 and 0.01818.
And the first 10 partial sums of the series 1.00000, 1.33333, 1.50000, 1.60000, 1.66667, 1.71429, 1.75000, 1.77778, 1.80000 and 1.81818.
The graph of the sequence of terms and the sequence of partial sums is shown below in Figure 1.
From the graph and the table, it is observed that the plotted points of the partial sums are closer to 2 and the terms of the sequence closer to zero.
Therefore, the series is convergent.
Obtain the sum of the series.
That is, to compute the values of
∑n=2∞an=∑n=2∞2n2−n
.
∑n=s∞2n2−n=limn→∞(∑k=2n2k(k−1))
(1)
Obtain the values of the series
∑k=1n2k(k−1)
.
The partial fraction of
2k(k−1)
is as follows,
2k(k−1)=a0k+a1k−1
(2)
Multiply by
k(k−1)
on both sides.
2⋅k(k−1)k(k−1)=a0⋅k(k−1)k+a1⋅k(k−1)k−1
2=a0(k−1)+a1k
(3)
Substitute 0 for k in equation (3),
2=a0(0−1)+a1⋅02=a0(−1)−2=a0
Thus, the value of
a0=−2
.
Substitute 1 for k in equation (3),
2=a0(1−1)+a1⋅12=a0(0)+a12=a1
Thus, the value of
a1=2
.
Substitute
−2 for a0 and 2 for a1
in equation (2),
2k(k−1)=−2k+2k−1=2k−1−2k
Therefore, the series becomes,
∑k=2n2k(k−1)=∑k=2n(2k−1−2k)=(22−1−22)+(23−1−23)+(24−1−24)+⋯+(2n−1−2n)=(21−22)+(22−23)+(23−24)+⋯+(2n−1−2n)=2−2n
.
Substitute
2−2n
for
∑k=2n2k(k−1)
in equation (1),
∑n=s∞2n2−n=limn→∞(2−2n)=limn→∞(2)−limn→∞(2n)=2−2⋅0=2
Thus, the series sum is 2.
Therefore, the series convergent to the sum is 2.