. The series is divergent. To see this, first observe that the series ∑ 1/kn for n = 1 to ∞ is divergent for any integer k ≥ 2.
Now, if we pick a large integer for k, say k > 100, then for nearly all integers n it will be true that 1 > cos(n) > 1/k. Therefore, since ∑ 1/kn is divergent, ∑ cos(n)/n must also be divergent The *summation* is divergent, but the individual terms converge to the number 0.<span>by comparison test since cosn/n <= 1/n is convergent
and 1/n is divergent by harmonic series
so the series is conditionally converget </span>
Answer:
c)The proof writer mentally assumed the conclusion. He wrote "suppose n is an arbitrary integer", but was really thinking "suppose n is an arbitrary integer, and suppose that for this n, there exists an integer k that satisfies n < k < n+2." Under those assumptions, it follows indeed that k must be n + 1, which justifies the word "therefore": but of course assuming the conclusion destroyed the validity of the proof.
Step-by-step explanation:
when we claim something as a hypothesis we can only conclude with therefore at the end of the proof. so assuming the conclusion nulify the proof from the beginning
The area of a trapezoid can be computed as:
A=(b1+b2)h/2, where b1 and b2 ar the bases, h is the height.
With the provided numerical values:
50=(3+7)h/2
50=10h/2=5h
Answer: h=10inches
Answer:
B) Vertex = (-5, 2)
Step-by-step explanation:
Vertex can be defined as defined as the point where two lines meet and form a particular angle with each other. It can also be described as the point where a line changes its direction
We can find vertex by two methods:
<h3>1) GRAPH</h3>
We can see in the graph attached below that the line changes its direction at point (-5, 2). So the vertex is (-5,2)
<h3>2) FORMULA</h3>
General form of equation of mod function is given by
y = a |x - h| + k
where Vertex = (h, k)
The given equation is
h(x) = 3 |x + 5| + 2
h(x) = 3 |x - (-5)| + 2
where h = -5 and k = 2
So the vertex is:
Vertex = (-5, 2)
Answer:
The last one is B.
Step-by-step explanation:
