Answer:
Take the integral
∫
∞
1
x
e
−
d
, which is finite, and note that it bounds
∑
n
=
2
. Therefore it is convergent, so
is as well.
Explanation:
The formal statement of the integral test states that if
f
∈
[
0
,
)
→
R
a monotone decreasing function which is non-negative. Then the sum
(
is convergent if and only if
sup
N
>
is finite. (Tau, Terence. Analysis I, second edition. Hindustan book agency. 2009).
This statement may seem a bit technical, but the idea is the following. Taking in this case the function
, we note that for
, this function is decreasing. We can see this by taking the derivative.
'
<
, since
, so
and
.
Due to this, we note that for any
≥
such that
≤
we have
. Therefore
+
∣
_
using integration by parts and that
lim
Since
, we have
3
. Since
, the series
increases as
increases. Since it is bounded by
, it must converge. Therefore
converges.
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Answers & Comments
Answer:
Take the integral
∫
∞
1
x
e
−
x
d
x
, which is finite, and note that it bounds
∞
∑
n
=
2
n
e
−
n
. Therefore it is convergent, so
∞
∑
n
=
1
n
e
−
n
is as well.
Explanation:
The formal statement of the integral test states that if
f
∈
[
0
,
∞
)
→
R
a monotone decreasing function which is non-negative. Then the sum
∞
∑
n
=
0
f
(
n
)
is convergent if and only if
sup
N
>
0
∫
N
0
f
(
x
)
d
x
is finite. (Tau, Terence. Analysis I, second edition. Hindustan book agency. 2009).
This statement may seem a bit technical, but the idea is the following. Taking in this case the function
f
(
x
)
=
x
e
−
x
, we note that for
x
>
1
, this function is decreasing. We can see this by taking the derivative.
f
'
(
x
)
=
e
−
x
−
x
e
−
x
=
(
1
−
x
)
e
−
x
<
0
, since
x
>
1
, so
(
1
−
x
)
<
0
and
e
−
x
>
0
.
Due to this, we note that for any
n
∈
N
≥
2
and
x
∈
[
1
,
∞
)
such that
x
≤
n
we have
f
(
x
)
≥
f
(
n
)
. Therefore
∫
n
n
−
1
f
(
x
)
d
x
≥
∫
n
n
−
1
f
(
n
)
d
x
=
f
(
n
)
, so
N
∑
n
=
1
f
(
n
)
≤
f
(
1
)
+
N
∑
n
=
2
∫
n
n
−
1
f
(
x
)
d
x
=
f
(
1
)
+
∫
N
1
f
(
x
)
d
x
.
∫
∞
1
f
(
x
)
d
x
=
∫
∞
1
x
e
−
x
d
x
=
−
∫
∞
x
=
1
x
d
e
−
x
=
−
x
e
−
x
∣
∞
1
+
∫
∞
1
e
−
x
d
x
=
−
x
e
−
x
−
e
−
x
∣
∞
_
1
=
2
e
using integration by parts and that
lim
x
→
∞
e
−
x
=
lim
x
→
∞
x
e
−
x
=
0
.
Since
f
(
x
)
≥
0
, we have
e
2
=
∫
∞
1
f
(
x
)
d
x
≥
∫
N
1
f
(
x
)
d
x
, so
N
∑
n
=
1
f
(
n
)
≤
f
(
1
)
+
2
e
=
3
e
. Since
f
(
n
)
≥
0
, the series
N
∑
n
=
1
f
(
n
)
increases as
N
increases. Since it is bounded by
3
e
, it must converge. Therefore
∞
∑
n
=
1
f
(
n
)
converges.