Terms (with the same numerator) in the series versus a single term in the series forĪll of the others. This calculation actually computes a pair of Series' equality with π/2 was presented by theģBlue1Brown youtube channel in 2018. The final series is the Wallis formula for π/2. Is constructed as follows with k starting at 0: The fifth series is a little less obvious as far as constructing the terms. We could not set some limit- it doesn't have Some value right here- and I haven't rigorouslyĭefined how we define a limit- but if this was not true, if Limit as n approaches infinity didn't go to N as n approaches infinity is 0, then we can say that a Positive and negative 1, but this denominatorĪnd bigger and bigger. Infinity, the numerator is going to oscillate between And I haven't rigorouslyĭefined it yet, but you can conceptualize Limited right over here to positive integers Is equivalent to asking what is the limit of The limit of a sub n as n approaches infinity
Sequence calculator convergence plus#
So we could write thisĪs equaling negative 1 to the n plus 1 power over n. When n is equal to 1, you haveġ times negative 1 squared, which is just 1, and it'll Negative 1 to the n, then this one would be negativeĪnd this would be positive. We start with a positive, thenĪ negative, positive, negative. With a sub n equaling- what does it equal? Well, if we ignore So we can define this sequenceĪs a sub n where n starts at 1 and goes to infinity Of a sub n as n approaches infinity? Well, let's think about if weĬan define a sub n explicitly. That is, what is the limit- let me do this in a new color. Which would make us askĪ very natural question- what happens to a sub nĪs n approaches infinity? Or another way of saying They kind of jump around, but they seem to be gettingĬloser and closer and closer to 0. And then when n isĮqual to 5, a sub n is equal to positive 1/5, which N is equal to negative 1/4, which is right about there. Sub n is equal to 1/3, which is right about there. When n is equal to 1, a sub n is equal to 1. Vertical and horizontal axes at the same scale, just so that Horizontal axis where I'm going to plot our n's. And it just keeps going onĪnd on and on like this. infinity + 1 = infinity, infinity^2 = infinity / 8, basically infinity can't be dealt with in algebra, (although L'Hopitol's rule lets us manage it), and shouldn't be thought of as a "number" in the common sense of the word. And remember, infinity is MUCH larger than 36, and this function will continue to leave potential values in the dust, never looking back, indefinitely.Īlso, just in case it isn't clear, infinity doesn't follow many of the properties that numbers follow. We might claim that it goes to 34359738368, a very big number equal to 2^35, but we 2^x, being fickle to the point of cruelty, will leave this value (already large) for a notably larger one as x goes to 36. What number does 2^x go to? (It diverges) If we were to investigate sin(x)/x, it would converge at 0, because the dividing by x heads to 0, and the +/- 1 can't stop it's approach.Ī similar resistance to staying mostly still can be found in equations that diverge as their inputs approach infinity. They don't head to infinity, and they don't converge. But, we know that they will always fluctuate. We know they will never output anything greater than 1, or less than -1, we are even able to compute them for any real number. One of the main things a function has to do to approach a number is to start to stabilize. Not all functions approach a number as their input approaches infinity. So we have an indeterminate form when we have a base approaching 1 and exponent approaching infinity, but not when we have a base that EQUALS 1 and exponent approaching infinity. We no longer have an infinitesimal increment away from 1 that can be overpowered by the increase of the exponent. The expression 1^b is always 1, no matter how large or small the exponent. HOWEVER, if a is not some function that approaches 1, but is actually the number 1, then we no longer have an indeterminate form. So that is why we say 1^∞ is an indeterminate form. If we find that a approaches 1 and b approaches infinity, we have an indeterminate form, because we can't tell without further analysis whether the forces attracting a toward 1 (making the expression approach 1) are overpowered by the forces moving b toward infinity (making the expression approach infinity or zero, depending on whether a is slightly greater than or less than 1). Suppose we want to know the limit of a^b as x goes to infinity, where a and b are both functions of x.
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