actual analysis – Test $ sum_ {k-1} ^ { infty} frac {1} {k + a} sqrt { frac {a} {k}}

$ forall a ge 0 $,Try out $ sum_ {k-1} ^ { infty} frac {1} {k + a} sqrt { frac {a} {k}} < pi $

The book hints at how to convert the sum to integration:$ sum_ {k-1} ^ { infty} frac {1} {(k + a) sqrt {k}} < int_ {0} ^ { infty} frac {dx} {(x + a) sqrt {x}} $, then I can use integration. But I can't prove it.

The test explains the algebraic integer in $ mathbb {Q} ( sqrt {d}) $

I have a question about the test in the book Elements of Number Theory (Stillwell). The idea is to test what number an algebraic integer can be in $ mathbb {Q} ( sqrt {d}) $.
My question is:
Why is this $ 2b $ under test (red color in the image)?

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asymptotic: why does $ frac {n ^ 3} {2 ^ { Omega ( sqrt { log n})}} $ refute the lower limit $ O (n ^ {3- delta}) $?

I have a simple question:

It is conjectured that the shortest path of all pairs (APSP) does not have $ O (n ^ {3- delta)} $time algorithm for any $ delta> 0 $ by SETH.

also

there is a result that says that APSP can be resolved in time $ frac {n ^ 3} {2 ^ { Omega ( sqrt { log n})}} $ by Ryan Williams.

But, this improvement does not refute the guesswork.

So what I did is this: I compare between $ lim_ {n -> infty} frac {( frac {n ^ 3} {2 ^ { sqrt { log n}}})} {n ^ {3- delta}} = 0 $ So, $ frac {n ^ 3} {2 ^ { Omega ( sqrt { log n})}} $ is better than the other, so why doesn't that mean we refute the conjecture?

When I have this function: $ frac {n ^ 3} {2 ^ { Omega ( sqrt { log n})}} $I did not know how to compare it with others, since Big Omega is only a part, how to compare in general with other functions when you have this?

Thanks in advance!

Why does the solution set of the equation $ sqrt {x-1} (x ^ 2-1) $ not include $ -1 $?

Why the solution set of the equation? $ sqrt {x-1} (x ^ 2-1) $ does not include $ -1 $. If I drink $ x = -1 $ I get $ 2i times 0 $ which is the same $ 0 $ and therefore $ -1 $ the solution should also be considered (along with $ + 1 $)

calculation – Evaluate $ lim_ {n rightarrow infty} sqrt {2 ^ n} int_a ^ b cos ^ n (x + frac { pi} {4}) dx $

Evaluate
$ lim_ {n rightarrow infty} sqrt {2 ^ n} int_a ^ b cos ^ n (x + frac { pi} {4}) dx $
Knowing that the integration set $ (a, b) $ is included in $ (0, frac { pi} {2}) $

Now, my attempt was to first notice that the function that we have to integrate is:$ cos ^ n (x + frac { pi} {4}) $ and I was thinking that I can use this trigonometric identity:$ cos (a + b) = cos (a) cos (b) + sin (a) sin (b) $, after which I can apply the reduction formula to $ cos ^ n (x) $. Is this the way to start?

What is the polar form of $ z = frac { sqrt (3) + i} { sqrt (3i)} $?

How to write the complex number $$ z = frac { sqrt (3) + i} { sqrt (3i)} $$ in polar form?

Upper limit $ int_0 ^ {+ infty} exp left (-a (-b + sqrt {b ^ 2 + cx}) ^ 2 right) dx $

Is there an upper limit depending on $ to $, $ b $ Y $ c $ (what positive constants) for the next intergral
begin {align}
int_0 ^ {+ infty} exp left (-a (-b + sqrt {b ^ 2 + cx}) ^ 2 right) dx.
end {align}

Can a real form (1,1) $ phi $ be represented by $ u sqrt {-1} partial bar { partial} u $ in the Kahler variety?

Let M be a 2-dimensional dimension (complex dimension) K "{a} ler and $ phi $ be a real $ (1,1) $ to form. Is it possible that there is a function? $ u $ such that $ phi = u sqrt {-1} partial bar { partial} u $

calculation – Evaluate $ lim limits_ {x a 0, x> 0} frac { int limits_0 ^ {sinx} sqrt {tant} dt} { int limits_0 ^ {tanx} sqrt {synt} dt} $

Solve: $$ l = lim limits_ {x to 0, x> 0} frac { int limits_0 ^ {sinx} sqrt {tant} dt} { int limits_0 ^ {tanx} sqrt {synt} dt} $$
So the main problem I have is evaluating $ int limits_0 ^ {tanx} sqrt {synt} dt $ I was looking to see exactly how to deal with this integral, but it seems that this integral requires college level skills (which I don't possess). This problem is from a high school textbook, so I thought there is some trick that can be applied without having to know about college. The main problem is with this integral because I have managed to solve the first one. Also, it looks like the limit itself will be a problem once you deal with the second integral (what I've tried to do with the second integral is substitute u =$ sqrt {synt} $ followed by $ sinx = u ^ 2 $ => cosxdx = 2u du => $ dx = frac {2u} { sqrt {1-u ^ 4}} du $ but I'm still getting stuck).

real analysis: where am I wrong when integrating $ int_ {0} ^ {1} { sqrt x} e ^ x dx $ per substitution?

$$
int_ {0} ^ {1} { sqrt x} e ^ x , dx
$$

My attempt:

I tried the replacement $$ { sqrt x} = t $$ and then I got $$ 2 int_ {0} ^ {1} te ^ { t ^ 2} , dt $$ So again I replaced $$ { t ^ 2} = u $$ and eventually $$ 2 int_ {0} ^ {1} frac { sqrt u} {2 { sqrt u}} e ^ u , du $$