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Convergence of series
Divergence, ratio, comparison and integral tests, and the harmonic series.
An infinite series converges when its partial sums approach a limit. The terms going to zero is necessary but not enough. The harmonic series 1 + ½ + ⅓ + … diverges even though its terms vanish. The ratio test settles most series with factorials or powers; comparison and the integral test handle the p-series family.
İşlənmiş nümunə · does 1/n^2 for n = 1 to oo converge
Addım-addım
- \sum_{n=1}^{\infty} \frac{1}{n^{2}}
Does this series converge?
- \lim_{n\to\infty} \frac{1}{n^{2}} = 0
Divergence test: if the terms do not go to 0 the series cannot converge.
- \lim \left|\frac{a_{k+1}}{a_k}\right| = \lim \left|{\frac{n^{2}}{\left(n + 1\right)^{2}}}\right| = 1
Ratio test: < 1 converges, > 1 diverges, = 1 says nothing.
- = \frac{\pi^{2}}{6}
It converges, and the CAS knows the sum.
Cavabı göstər
İndi siz Problem seçin, ya da özünü yazın ya da çəkin. Hər addım, bir rəsm, cavab siz soruşana qədər gizli qalır.
Burada istifadə edilən simvollar
Tam təsviri, şəkli və hər hərfin nə demək olduğunu görmək üçün hər hansı simvolu vurun.
Növbəti: Convergence of series
- Does this series converge?
- Divergence test: if the terms do not go to 0 the series cannot converge.
- Ratio test: < 1 converges, > 1 diverges, = 1 says nothing.
- It converges, and the CAS knows the sum.
İnsanlar soruşurlar
What is the ε, δ definition actually saying?
That you can make the output as close to the limit as anyone demands (within ε) by keeping the input close enough (within δ). It replaces "approaches" with a challenge-and-response that can be checked.
Why does the harmonic series diverge when its terms go to zero?
Because the terms shrink too slowly: group them as 1/3 + 1/4 > 1/2, 1/5 + … + 1/8 > 1/2, and so on: infinitely many halves.
Daha çox Real Analysis
Sequences and their limitsImproper integralsTaylor approximation and errorFunctions of a complex variableContinuity and differentiability, rigorouslySequences of functions and uniform convergence