Question

I need help..

Answer 1

unblock me por favor

Answer 2

what type of math iz that?

Answer 3

She's on that Einstein level.

Answer 4

Hi

Answer 5

Is anyone a girl here?? I need some help

Answer 6

Apply the limit divergence test

Answer 7

As gets closer to 400, the given statement represents a limit. Let's dissect it step-by-step:

1. The limit is denoted by the symbol "lim".
2. The variable is getting close to 400.
3. M k=1 1 — k is the expression inside the limit.
In n

Understanding the terms involved is necessary to evaluate this expression:
- The series' total number of terms is denoted by the letter M.
- The notation k=1 in the sigma notation denotes that we begin with k = 1 and proceed up to M.
- The term of the series is represented by the phrase 1/k.
- The logarithm with base e is known as the natural logarithm of n, abbreviated as In n.

Let's now make the phrase simpler:
Using the sigma notation, combine the terms as follows:
1/1 + 1/2 + 1/3 + ... + 1/M

2. Calculate each term's natural logarithm as follows: In 1 + In 2 + In 3 +... + In M

It is unclear at this time what value of n we should employ.

Answer 8

i have no idea how to do that

Answer 9

i wish i did

Answer 10

To solve the student's question, we can follow a similar approach as before. Let's break it down step by step:

Step 1: Rewrite the expression using the limit notation:
γ
=
lim
⁡
n
→
∞
(
∑
k
=
1
n
1
k
−
ln
⁡
n
)
γ=
n→∞
lim
​
(
k=1
∑
n
​

k
1
​
−lnn)

Step 2: Rewrite the summation as an infinite series:
γ
=
lim
⁡
n
→
∞
(
∑
k
=
1
∞
1
k
−
ln
⁡
n
)
γ=
n→∞
lim
​
(
k=1
∑
∞
​

k
1
​
−lnn)

Step 3: Simplify the expression inside the limit:
γ
=
∑
k
=
1
∞
1
k
−
lim
⁡
n
→
∞
ln
⁡
n
γ=
k=1
∑
∞
​

k
1
​
−
n→∞
lim
​
lnn

Step 4: Evaluate the limit of
ln
⁡
n
lnn as
n
n approaches infinity:
lim
⁡
n
→
∞
ln
⁡
n
=
+
∞
n→∞
lim
​
lnn=+∞

Step 5: Rewrite the expression with the limit of
ln
⁡
n
lnn:
γ
=
∑
k
=
1
∞
1
k
−
(
+
∞
)
γ=
k=1
∑
∞
​

k
1
​
−(+∞)

Step 6: Since the sum of the series
∑
k
=
1
∞
1
k
k=1
∑
∞
​

k
1
​
is known to be divergent (it goes to infinity), we can say that
∑
k
=
1
∞
1
k
−
(
+
∞
)
k=1
∑
∞
​

k
1
​
−(+∞) is also divergent.